I've already discussed a 'third' method to solve the equation x³ = 9x + 12, but how about a 'fourth' method? It is also possible to solve depressed cubic equations with a single real root by means of a _hyperbolic substitution._ First rewrite the equation as (1) x³ − 9x = 12 and then substitute (2) x = r·cosh t which gives (3) r³·cosh³t − 9r·cosh t = 12 The idea is to take advantage of the triple argument identity (4) cosh 3t = 4·cosh³t − 3·cosh t by turning the left hand side of (3) into 4·cosh³t − 3·cosh t by selecting a suitable value of r. In order to do this, we first multiply both sides of (3) by 4/r³ which gives (5) 4·cosh³t − (36/r²)·cosh t = 48/r³ Now, we only need to choose a value for r in (5) such that 36/r² = 3 which implies r² = 12, so we can choose (6) r = 2√3 With this value of r we have 48/r³ = 48/(24√3) = 2/√3 = ⅔√3 so (5) becomes (7) 4·cosh³t − 3·cosh t = ⅔√3 and because of identity (4) this can be written as (8) cosh 3t = ⅔√3 so we have (9) 3t = arcosh(⅔√3) Now, arcosh x = ln(x + √(x² − 1)) so arcosh(⅔√3) = ln(⅔√3 + √(⁴⁄₃ − 1)) = ln(⅔√3 + ⅓√3) = ln(√3) = ½·ln(3) so we have 3t = ½·ln(3) and therefore (10) t = ⅙·ln(3) and since we substituted x = r·cosh t where r = 2√3 it follows that the real root of (2) is (11) x = 2√3·cosh(⅙·ln(3)) To convert this into algebraic form, we note that √3 = 3¹ᐟ², ⅙·ln(3) = ln(3¹ᐟ⁶), cosh t = ½(eᵗ + e⁻ᵗ) so we have x = 2√3·cosh(⅙·ln(3)) = 2·3¹ᐟ²·(½)·(3¹ᐟ⁶ + 3⁻¹ᐟ⁶) = 3¹ᐟ² ⁺ ¹ᐟ⁶ + 3¹ᐟ² ⁻ ¹ᐟ⁶ = 3²ᐟ³ + 3¹ᐟ³ = ∛9 + ∛3 as expected. It is also possible to obtain the complex roots of (1) by noting that the complex hyperbolic cosine is a periodic function with a period 2πi so we actually have 3t = ½·ln(3) + 2kπi and therefore (12) t = ⅙·ln(3) + ⅔·k·πi for any integer k and where all three roots can be obtained by selecting any three consecutive integer values for k. The real root is obtained with e.g. k = 0 and the two complex roots can be obtained by selecting e.g. k = 1 and k = −1. Since exp(⅔·πi) = −½ + i·½√3 = ω₁ and exp(−⅔·πi) = −½ − i·½√3 = ω₂ are the complex cube roots of unity, we therefore get x₁ = 2√3·cosh(⅙·ln(3)) = ∛9 + ∛3 x₂ = 2√3·cosh(⅙·ln(3) + ⅔·πi) = ω₁∛9 + ω₂∛3 = −½(∛9 + ∛3) + i·½√3·(∛9 − ∛3) x₃ = 2√3·cosh(⅙·ln(3) − ⅔·πi) = ω₂∛9 + ω₁∛3 = −½(∛9 + ∛3) − i·½√3·(∛9 − ∛3) which is of course in agreement with the solutions obtained by other methods.
@@SyberMath I wonder if you did see my 'third' method which involves transforming your cubic into 3(x + 1)³ = (x + 3)³ It seems that some of my RUclips comments are only visible in my own account, and not visible for anyone else. Very annoying.
0:30 Rather than fiddling around with minus signs, it is easier to keep the plus sign: x^3 = 9x + 12 And make the Ansatz x = u + v. We know from the binomic formulae that (u + v)^3 = u^3 + 3u^2v + 3uv^2 + v^3 = u^3 + 3uv(u + v) + v^3 Now replace x by u + v in the cubic equation: (u + v)^3 = 9(u + v) + 12 Thus u^3 + 3uv(u + v) + v^3 = 9(u + v) + 12 By comparison of cofficients, we see that this system is satisfied if u^3 + v^3 = 12 and 3uv = 9 or uv = 3 (uv)^3 = 27 u^3 * v^3 = 27 We know the sum (12) and product (27) of u^3 and v^3. According to Vieta, we can thus compute u^3 and v^3 as solutions z1, z2 of the quadratic z^2 - 12z + 27 = 0 z^2 - 12z + 36 = 9 (z - 6)^2 = 9 z - 6 = +- 3 z1 = 6 + 3 = 9 z2 = 6 - 3 = 3 So the first (and only real) solution of our cubic is x = u + v = cuberoot(3) + cuberoot(9) as in the video.
x^3 = 9x + 12 Choose the value of t such that x = t - k/t. k is, at this point, an unknown constant, but we will pick a value for it later. Since x = t - k/t, we can calculate x^3 = (t - k/t)^3 = t^3 - 3kt + 3k^2 /t - k^3 / t^3 On the other hand, we can see rather easily that 9x + 12 = 9t - 9k/t + 12 So, we have t^3 - 3kt + 3k^2 / t - k^3 / t^3 = 9t - 9k/t + 12. Moving things around, we get t^3 - (3k + 9)t + (3k^2 + 9k) / t - k^3 / t^3 = 12. It sure would be nice if 3k + 9 = 0 and also 3k^2 + 9k = 0. Wait... that happens when k = -3! I said we'd pick a value for k later, we'll pick it now. k = -3. So, x = t + 3/t, and we now have t^3 + 27 / t^3 = 12. Let u = t^3 and we end up with u + 27 / u = 12. Multiply by u, we get u^2 + 27 = 12u, or u^2 - 12u + 27 = 0. Take your preferred way of solving a quadratic, and we get u = 3 or u = 9. Since u = t^3, we now have t = cbrt(3) or cbrt(9). And since x = t + 3/t, we get x = cbrt(3) + 3/cbrt(3) or x = cbrt(9) + 3/cbrt(9). It turns out that those statements are equivalent, so we only have one solution, but it is the same solution you got.
The cubic formula I know means substituting x = a(u + 1/u). Then setting a so that the (u + 1/u) term is eliminated. This still results in a quadratic in u^3, and the same values in the solution.
I think a+b should give complete set of x, a^3=9 so it will have 1 real and 2 imaginary roots (same for b) and therefore x will only have 1 real value.. Correct me if I am wrong
Well that is not entirely true. If there are one real and two complex solutions for a (and likewise for b) then the equation indeed has one real and two (conjugate) complex solutions. However, if the system in a and b has no real solutions, then, perhaps somewhat surpringly for you, the equation will have three real roots. See my main comment on this video where I explain this in detail.
By inspection we know one root is between 3 and 4, so we will use fixed-point iteration x←∛(9x+12), starting with x=3.5, to get x=3.522333393359312... Then synthetically divide x^3-9x-12=0 by (x-3.522333393359312...) to get the b and c coefficients of a quadratic equation, x^2+bx+c=0, that may be solved using the quadratic formula for the remaining two roots, x= [-b±√(b^2-4c)]/2.
@@farhansadik5423 Suppose we have a cubic of the form x^3 = 3ax + b. (The 3 will just make things nicer if I put it here). Suppose we have a t such that x = t + a/t. Substituting that in, we get t^3 + 3at + 3a^2 / t + a^3 / t^3 = 3at + 3a^2 / t + b. But the 3at and 3a^2 / t terms cancel out, so we end up with t^3 + a^3 / t^3 = b. Let u = t^3, we have u + a^3 / u = b. Multiply by u, we get u^2 + a^3 = bu. That's a quadratic in u, so we can solve it. Then t is the cube root of that, and we can get x by the equation x = t + a/t.
@@chaosredefined3834 man you're an absolute legend! I think this is like called cardano's method to solve depressed cubics or something, but your explanations is superb! May god bless you!
Wait, why do they(3at and 3a^2/t) cancel out again? Sorry, it's 3 am and my brains little slow. Thanks for the explanation though Edit:Nvm, I understand it, thats so cool!
Since you asked for a third method to solve the equation x³ = 9x + 12 how about the following method. I don't think you or anyone else here has seen this method before, but nonetheless it was partially inspired by what you did earlier with some other cubic equations. The idea is to convert this equation into an equation of the form p(x + r₁)³ = q(x + r₂)³ because then we only need to take the cube roots from both sides to get a linear equation in x and then we can at least find the (only) real root of the equation. In case you are wondering why I chose p, q , r₁, r₂ and not just p, q, r, s or something similar, this will become clear in due course. If we expand both sides and then bring all terms from the right hand side over to the left hand side we can write this equation as (p − q)x³ + 3(pr₁ − qr₂)x² + 3(pr₁² − qr₂²) + (pr₁³ − qr₂³) = 0 and comparing coefficients with x³ − 9x − 12 = 0 we see that p, q, r₁, r₂ must satisfy (1) p − q = 1 (2) pr₁ − qr₂ = 0 (3) pr₁² − qr₂² = −3 (4) pr₁³ − qr₂³ = −12 Now this is not a particularly easy system to solve. I won't go through all the intermediate steps since that is just elementary algebra, but we can proceed as follows. First we multiply the left hand sides of (1) and (3) and then we subtract the square of the left hand side of (2) which gives (5) (p − q)(pr₁² − qr₂²) − (pr₁ − qr₂)² = −pq(r₁ − r₂)² Then we multiply the left hand sides of (1) and (4) and subtract the product of the left hand sides of (2) and (3) which gives (6) (p − q )(pr₁³ − qr₂³) − (pr₁ − qr₂)(pr₁² − qr₂²) = −pq(r₁ − r₂)²(r₁ + r₂) And finally we multiply the left hand sides of (2) and (4) and then subtract the square of the left hand side of (3) which gives (7) (pr₁ − qr₂)(pr₁³ − qr₂³) − (pr₁² − qr₂²)² = −pqr₁r₂(r₁ − r₂)² Substituting (1), (2), (3), (4) in (5), (6), (7) we find that we have (8) −pq(r₁ − r₂)² = −3 (9) −pq(r₁ − r₂)²(r₁ + r₂) = −12 (10) −pqr₁r₂(r₁ − r₂)² = −9 Dividing (9) by (8) we have (11) r₁ + r₂ = 4 and dividing (10) by (8) we have (12) r₁r₂ = 3 which means that r₁ and r₂ are the roots of the quadratic equation (13) r² − 4r + 3 = 0 which factors as (r − 1)(r − 3) = 0 so we have r₁ = 1, r₂ = 3 or vice versa. Substituting r₁ = 1, r₂ = 3 in (2) we have p − 3q = 0 so p = 3q and substituting this in (1) we have q = ¹⁄₂ and so p = ³⁄₂. So, we have a solution (p, q, r₁, r₂) = (³⁄₂, ¹⁄₂, 1, 3) for our system (1), (2), (3), (4). Of course, we can also choose r₁ = 3, r₂ = 1 and then we find p = −¹⁄₂, q = −³⁄₂ so (p, q, r₁, r₂) = (−¹⁄₂, −³⁄₂, 3, 1) is another solution of the same system. For (p, q, r₁, r₂) = (³⁄₂, ¹⁄₂, 1, 3) the equation p(x + r₁)³ = q(x + r₂)³ becomes ³⁄₂(x + 1)³ = ¹⁄₂(x + 3)³ and multiplying both sides by 2 to get rid of the fractions this gives 3(x + 1)³ = (x + 3)³ which is indeed equivalent with our original equation x³ = 9x + 12 as can be verified by expanding both sides. Rewriting this equation as (∛3·x + ∛3)³ = (x + 3)³ we can see that for the real root of this equation we have ∛3·x + ∛3 = x + 3 (∛3 − 1)x = 3 − ∛3 x = (3 − ∛3)/(∛3 − 1) To rationalize the denominator ∛3 − 1 we let a = ∛3, b = 1 which gives 1/(∛3 − 1) = 1/(a − b) = (a² + ab + b²)/(a³ − b³) = (∛9 + ∛3 + 1)/2 and multiplying this by the numerator 3 − ∛3 we have x = ½(3 − ∛3)(∛9 + ∛3 + 1) = ½(3∛9 + 3∛3 + 3 − 3 − ∛9 − ∛3) = ½(2∛9 + 2∛3) = ∛9 + ∛3 which is of course the same real root you found.
@@SyberMath Thank you. In case you are wondering how I figured this out: I started by doing some reverse engineering on your cubic equation x³ = 9x + 12 Of course I already knew the real root x = ∛9 + ∛3. The interesting thing about this expression is that (∛3)² = ∛9. Why is this interesting? Well, depressed cubic equations of course have a real solution which is the sum of two cube roots. With non-depressed cubic equations there is an additional constant which is the same for all three roots, because to turn a non-depressed cubic into a depressed cubic the roots are simply shifted in such a way that the sum of the roots of the equation in the new variable is zero (which causes the quadratic term to drop out). Now, we can of course also reverse this process, i.e. shifting the roots of a depressed cubic to turn it into a non-depressed cubic. To create a cubic in a new variable z from your cubic in x where each root is incremented by 1 we would need to have z = x + 1 and then the real root of this new cubic in z will obviously be z = ∛9 + ∛3 + 1 But why did I choose to add 1 to the roots of the cubic in x for my new cubic in z? Well, exactly because of ∛9 = (∛3)². The three terms now form a geometric series. If we let ∛3 = a then we have z = a² + a + 1 = (a³ − 1)/(a − 1) = (3 − 1)/(∛3 − 1) = 2/(∛3 − 1) From this we can work back to a cubic equation in z where both sides are a cube: (∛3 − 1)z = 2 ∛3·z − z = 2 ∛3·z = z + 2 3z³ = (z + 2)³ and since z = x + 1 this gives 3(x + 1)³ = (x + 3)³ as a cubic equation of the desired form which is equivalent with x³ = 9x + 12 as you can easily verify. So, I already knew exactly what transformation I needed to achieve. Then it was just a matter of setting up an equation p(x + r₁)³ = q(x + r₂)³ and expand and compare coefficients to get a system of equations in p, q, r₁, r₂. This system turned out to have two solutions (p, q, r₁, r₂) = (³⁄₂, ¹⁄₂, 1, 3) and (p, q, r₁, r₂) = (−¹⁄₂, −³⁄₂, 3, 1). With the first solution we get ³⁄₂(x + 1)³ = ¹⁄₂(x + 3)³ where we only need to multiply both sides by 2 to obtain the equation which resulted from reverse engineering your original cubic. With the second solution we get −¹⁄₂(x + 3)³ = −³⁄₂(x + 1)³ which is clearly the same equation with the left and right hand sides swapped and multiplied by −1. If we expand both sides of both these equations and bring all terms over to the left hand side then _both_ equations result in x³ − 9x − 12 = 0 which is exactly as it should be. So, the fact that the system in p, q, r₁, r₂ has two solutions and the minus signs of p and q with the second solution make perfect sense.
substitute ; x = y + 3/y y³ + 27/y³ - 12 = 0 moreover, substitute ; z = y³ z + 27/z - 12 = 0, z ≠ 0 z² - 12z + 27 = 0 ; z = 3, 9 either z = 3 or 9, we get same value x = ∛3 + ∛9 the other two solutions are complex, 2y = (-1 + i√3)*∛z and (-1 - i√3)*∛z
if you actually try to do the second method the systems of equations require you to solve the original cubic so the second method is useless, why even mention it
@SyberMath but why start with a plus b to being with..ypu have to agree it comes out of nowhere...unless you first rule out integers and rational as solutions. .sona plus b represents a rational plus irrational number..is that why orif not why..I hope you can PLEASE PLEASE ANSWER
@vladimirkaplun5774: See my main comment on this video for an entirely different method to solve cubics which has nothing to do with the method named after Cardano and which you have probably never seen before.
I've already discussed a 'third' method to solve the equation x³ = 9x + 12, but how about a 'fourth' method?
It is also possible to solve depressed cubic equations with a single real root by means of a _hyperbolic substitution._
First rewrite the equation as
(1) x³ − 9x = 12
and then substitute
(2) x = r·cosh t
which gives
(3) r³·cosh³t − 9r·cosh t = 12
The idea is to take advantage of the triple argument identity
(4) cosh 3t = 4·cosh³t − 3·cosh t
by turning the left hand side of (3) into 4·cosh³t − 3·cosh t by selecting a suitable value of r. In order to do this, we first multiply both sides of (3) by 4/r³ which gives
(5) 4·cosh³t − (36/r²)·cosh t = 48/r³
Now, we only need to choose a value for r in (5) such that 36/r² = 3 which implies r² = 12, so we can choose
(6) r = 2√3
With this value of r we have 48/r³ = 48/(24√3) = 2/√3 = ⅔√3 so (5) becomes
(7) 4·cosh³t − 3·cosh t = ⅔√3
and because of identity (4) this can be written as
(8) cosh 3t = ⅔√3
so we have
(9) 3t = arcosh(⅔√3)
Now, arcosh x = ln(x + √(x² − 1)) so arcosh(⅔√3) = ln(⅔√3 + √(⁴⁄₃ − 1)) = ln(⅔√3 + ⅓√3) = ln(√3) = ½·ln(3) so we have 3t = ½·ln(3) and therefore
(10) t = ⅙·ln(3)
and since we substituted x = r·cosh t where r = 2√3 it follows that the real root of (2) is
(11) x = 2√3·cosh(⅙·ln(3))
To convert this into algebraic form, we note that √3 = 3¹ᐟ², ⅙·ln(3) = ln(3¹ᐟ⁶), cosh t = ½(eᵗ + e⁻ᵗ) so we have
x = 2√3·cosh(⅙·ln(3)) = 2·3¹ᐟ²·(½)·(3¹ᐟ⁶ + 3⁻¹ᐟ⁶) = 3¹ᐟ² ⁺ ¹ᐟ⁶ + 3¹ᐟ² ⁻ ¹ᐟ⁶ = 3²ᐟ³ + 3¹ᐟ³ = ∛9 + ∛3
as expected.
It is also possible to obtain the complex roots of (1) by noting that the complex hyperbolic cosine is a periodic function with a period 2πi so we actually have 3t = ½·ln(3) + 2kπi and therefore
(12) t = ⅙·ln(3) + ⅔·k·πi
for any integer k and where all three roots can be obtained by selecting any three consecutive integer values for k. The real root is obtained with e.g. k = 0 and the two complex roots can be obtained by selecting e.g. k = 1 and k = −1. Since exp(⅔·πi) = −½ + i·½√3 = ω₁ and exp(−⅔·πi) = −½ − i·½√3 = ω₂ are the complex cube roots of unity, we therefore get
x₁ = 2√3·cosh(⅙·ln(3)) = ∛9 + ∛3
x₂ = 2√3·cosh(⅙·ln(3) + ⅔·πi) = ω₁∛9 + ω₂∛3 = −½(∛9 + ∛3) + i·½√3·(∛9 − ∛3)
x₃ = 2√3·cosh(⅙·ln(3) − ⅔·πi) = ω₂∛9 + ω₁∛3 = −½(∛9 + ∛3) − i·½√3·(∛9 − ∛3)
which is of course in agreement with the solutions obtained by other methods.
😍❤️
@@SyberMath I wonder if you did see my 'third' method which involves transforming your cubic into
3(x + 1)³ = (x + 3)³
It seems that some of my RUclips comments are only visible in my own account, and not visible for anyone else. Very annoying.
0:30 Rather than fiddling around with minus signs, it is easier to keep the plus sign:
x^3 = 9x + 12
And make the Ansatz x = u + v.
We know from the binomic formulae that
(u + v)^3
= u^3 + 3u^2v + 3uv^2 + v^3
= u^3 + 3uv(u + v) + v^3
Now replace x by u + v in the cubic equation:
(u + v)^3 = 9(u + v) + 12
Thus
u^3 + 3uv(u + v) + v^3 = 9(u + v) + 12
By comparison of cofficients, we see that this system is satisfied if
u^3 + v^3 = 12
and
3uv = 9 or
uv = 3
(uv)^3 = 27
u^3 * v^3 = 27
We know the sum (12) and product (27) of u^3 and v^3.
According to Vieta, we can thus compute u^3 and v^3 as solutions z1, z2 of the quadratic
z^2 - 12z + 27 = 0
z^2 - 12z + 36 = 9
(z - 6)^2 = 9
z - 6 = +- 3
z1 = 6 + 3 = 9
z2 = 6 - 3 = 3
So the first (and only real) solution of our cubic is
x = u + v = cuberoot(3) + cuberoot(9)
as in the video.
Nice!
x^3 = 9x + 12
Choose the value of t such that x = t - k/t. k is, at this point, an unknown constant, but we will pick a value for it later.
Since x = t - k/t, we can calculate x^3 = (t - k/t)^3 = t^3 - 3kt + 3k^2 /t - k^3 / t^3
On the other hand, we can see rather easily that 9x + 12 = 9t - 9k/t + 12
So, we have t^3 - 3kt + 3k^2 / t - k^3 / t^3 = 9t - 9k/t + 12. Moving things around, we get t^3 - (3k + 9)t + (3k^2 + 9k) / t - k^3 / t^3 = 12. It sure would be nice if 3k + 9 = 0 and also 3k^2 + 9k = 0. Wait... that happens when k = -3! I said we'd pick a value for k later, we'll pick it now. k = -3. So, x = t + 3/t, and we now have t^3 + 27 / t^3 = 12. Let u = t^3 and we end up with u + 27 / u = 12.
Multiply by u, we get u^2 + 27 = 12u, or u^2 - 12u + 27 = 0. Take your preferred way of solving a quadratic, and we get u = 3 or u = 9. Since u = t^3, we now have t = cbrt(3) or cbrt(9). And since x = t + 3/t, we get x = cbrt(3) + 3/cbrt(3) or x = cbrt(9) + 3/cbrt(9). It turns out that those statements are equivalent, so we only have one solution, but it is the same solution you got.
The cubic formula I know means substituting x = a(u + 1/u). Then setting a so that the (u + 1/u) term is eliminated.
This still results in a quadratic in u^3, and the same values in the solution.
Nice!
if we calculate the cbrt(9) and cbrt(3) as complex numbers, we will get the three factors (plus 3 extraneous solutions).
Yes, I can😊
I think a+b should give complete set of x, a^3=9 so it will have 1 real and 2 imaginary roots (same for b) and therefore x will only have 1 real value..
Correct me if I am wrong
Well that is not entirely true. If there are one real and two complex solutions for a (and likewise for b) then the equation indeed has one real and two (conjugate) complex solutions. However, if the system in a and b has no real solutions, then, perhaps somewhat surpringly for you, the equation will have three real roots. See my main comment on this video where I explain this in detail.
By inspection we know one root is between 3 and 4, so we will use fixed-point iteration x←∛(9x+12), starting with x=3.5, to get x=3.522333393359312...
Then synthetically divide x^3-9x-12=0 by (x-3.522333393359312...) to get the b and c coefficients of a quadratic equation, x^2+bx+c=0, that may be solved using the quadratic formula for the remaining two roots, x= [-b±√(b^2-4c)]/2.
Solve using brute force - Cardano.
Nice!
Thank you! 😍
Beautifull
Please consider the solution method through the substitution y = x + k/x for cubic eqn's in your next videos
woah, what is that method? What is it called or how is it used to solve cubics?
@@farhansadik5423 Suppose we have a cubic of the form x^3 = 3ax + b. (The 3 will just make things nicer if I put it here). Suppose we have a t such that x = t + a/t. Substituting that in, we get t^3 + 3at + 3a^2 / t + a^3 / t^3 = 3at + 3a^2 / t + b. But the 3at and 3a^2 / t terms cancel out, so we end up with t^3 + a^3 / t^3 = b. Let u = t^3, we have u + a^3 / u = b. Multiply by u, we get u^2 + a^3 = bu. That's a quadratic in u, so we can solve it. Then t is the cube root of that, and we can get x by the equation x = t + a/t.
@@farhansadik5423 I don't know the name of the method, but that's how it's used.
@@chaosredefined3834 man you're an absolute legend! I think this is like called cardano's method to solve depressed cubics or something, but your explanations is superb! May god bless you!
Wait, why do they(3at and 3a^2/t) cancel out again? Sorry, it's 3 am and my brains little slow. Thanks for the explanation though
Edit:Nvm, I understand it, thats so cool!
Pozdrawiam serdecznie i życzę miłego dnia
Thank you! Likewise
Since you asked for a third method to solve the equation
x³ = 9x + 12
how about the following method. I don't think you or anyone else here has seen this method before, but nonetheless it was partially inspired by what you did earlier with some other cubic equations. The idea is to convert this equation into an equation of the form
p(x + r₁)³ = q(x + r₂)³
because then we only need to take the cube roots from both sides to get a linear equation in x and then we can at least find the (only) real root of the equation. In case you are wondering why I chose p, q , r₁, r₂ and not just p, q, r, s or something similar, this will become clear in due course. If we expand both sides and then bring all terms from the right hand side over to the left hand side we can write this equation as
(p − q)x³ + 3(pr₁ − qr₂)x² + 3(pr₁² − qr₂²) + (pr₁³ − qr₂³) = 0
and comparing coefficients with
x³ − 9x − 12 = 0
we see that p, q, r₁, r₂ must satisfy
(1) p − q = 1
(2) pr₁ − qr₂ = 0
(3) pr₁² − qr₂² = −3
(4) pr₁³ − qr₂³ = −12
Now this is not a particularly easy system to solve. I won't go through all the intermediate steps since that is just elementary algebra, but we can proceed as follows. First we multiply the left hand sides of (1) and (3) and then we subtract the square of the left hand side of (2) which gives
(5) (p − q)(pr₁² − qr₂²) − (pr₁ − qr₂)² = −pq(r₁ − r₂)²
Then we multiply the left hand sides of (1) and (4) and subtract the product of the left hand sides of (2) and (3) which gives
(6) (p − q )(pr₁³ − qr₂³) − (pr₁ − qr₂)(pr₁² − qr₂²) = −pq(r₁ − r₂)²(r₁ + r₂)
And finally we multiply the left hand sides of (2) and (4) and then subtract the square of the left hand side of (3) which gives
(7) (pr₁ − qr₂)(pr₁³ − qr₂³) − (pr₁² − qr₂²)² = −pqr₁r₂(r₁ − r₂)²
Substituting (1), (2), (3), (4) in (5), (6), (7) we find that we have
(8) −pq(r₁ − r₂)² = −3
(9) −pq(r₁ − r₂)²(r₁ + r₂) = −12
(10) −pqr₁r₂(r₁ − r₂)² = −9
Dividing (9) by (8) we have
(11) r₁ + r₂ = 4
and dividing (10) by (8) we have
(12) r₁r₂ = 3
which means that r₁ and r₂ are the roots of the quadratic equation
(13) r² − 4r + 3 = 0
which factors as (r − 1)(r − 3) = 0 so we have r₁ = 1, r₂ = 3 or vice versa. Substituting r₁ = 1, r₂ = 3 in (2) we have p − 3q = 0 so p = 3q and substituting this in (1) we have q = ¹⁄₂ and so p = ³⁄₂. So, we have a solution (p, q, r₁, r₂) = (³⁄₂, ¹⁄₂, 1, 3) for our system (1), (2), (3), (4). Of course, we can also choose r₁ = 3, r₂ = 1 and then we find p = −¹⁄₂, q = −³⁄₂ so (p, q, r₁, r₂) = (−¹⁄₂, −³⁄₂, 3, 1) is another solution of the same system. For (p, q, r₁, r₂) = (³⁄₂, ¹⁄₂, 1, 3) the equation
p(x + r₁)³ = q(x + r₂)³
becomes
³⁄₂(x + 1)³ = ¹⁄₂(x + 3)³
and multiplying both sides by 2 to get rid of the fractions this gives
3(x + 1)³ = (x + 3)³
which is indeed equivalent with our original equation x³ = 9x + 12 as can be verified by expanding both sides. Rewriting this equation as
(∛3·x + ∛3)³ = (x + 3)³
we can see that for the real root of this equation we have
∛3·x + ∛3 = x + 3
(∛3 − 1)x = 3 − ∛3
x = (3 − ∛3)/(∛3 − 1)
To rationalize the denominator ∛3 − 1 we let a = ∛3, b = 1 which gives 1/(∛3 − 1) = 1/(a − b) = (a² + ab + b²)/(a³ − b³) = (∛9 + ∛3 + 1)/2 and multiplying this by the numerator 3 − ∛3 we have
x = ½(3 − ∛3)(∛9 + ∛3 + 1) = ½(3∛9 + 3∛3 + 3 − 3 − ∛9 − ∛3) = ½(2∛9 + 2∛3) = ∛9 + ∛3
which is of course the same real root you found.
Amazing! I always learn new things from you 😍❤️
@@SyberMath Thank you. In case you are wondering how I figured this out: I started by doing some reverse engineering on your cubic equation
x³ = 9x + 12
Of course I already knew the real root x = ∛9 + ∛3. The interesting thing about this expression is that (∛3)² = ∛9. Why is this interesting? Well, depressed cubic equations of course have a real solution which is the sum of two cube roots. With non-depressed cubic equations there is an additional constant which is the same for all three roots, because to turn a non-depressed cubic into a depressed cubic the roots are simply shifted in such a way that the sum of the roots of the equation in the new variable is zero (which causes the quadratic term to drop out).
Now, we can of course also reverse this process, i.e. shifting the roots of a depressed cubic to turn it into a non-depressed cubic. To create a cubic in a new variable z from your cubic in x where each root is incremented by 1 we would need to have
z = x + 1
and then the real root of this new cubic in z will obviously be
z = ∛9 + ∛3 + 1
But why did I choose to add 1 to the roots of the cubic in x for my new cubic in z? Well, exactly because of ∛9 = (∛3)². The three terms now form a geometric series. If we let ∛3 = a then we have
z = a² + a + 1 = (a³ − 1)/(a − 1) = (3 − 1)/(∛3 − 1) = 2/(∛3 − 1)
From this we can work back to a cubic equation in z where both sides are a cube:
(∛3 − 1)z = 2
∛3·z − z = 2
∛3·z = z + 2
3z³ = (z + 2)³
and since z = x + 1 this gives
3(x + 1)³ = (x + 3)³
as a cubic equation of the desired form which is equivalent with
x³ = 9x + 12
as you can easily verify. So, I already knew exactly what transformation I needed to achieve. Then it was just a matter of setting up an equation
p(x + r₁)³ = q(x + r₂)³
and expand and compare coefficients to get a system of equations in p, q, r₁, r₂. This system turned out to have two solutions (p, q, r₁, r₂) = (³⁄₂, ¹⁄₂, 1, 3) and (p, q, r₁, r₂) = (−¹⁄₂, −³⁄₂, 3, 1). With the first solution we get
³⁄₂(x + 1)³ = ¹⁄₂(x + 3)³
where we only need to multiply both sides by 2 to obtain the equation which resulted from reverse engineering your original cubic. With the second solution we get
−¹⁄₂(x + 3)³ = −³⁄₂(x + 1)³
which is clearly the same equation with the left and right hand sides swapped and multiplied by −1. If we expand both sides of both these equations and bring all terms over to the left hand side then _both_ equations result in
x³ − 9x − 12 = 0
which is exactly as it should be. So, the fact that the system in p, q, r₁, r₂ has two solutions and the minus signs of p and q with the second solution make perfect sense.
nice trick
substitute ; x = y + 3/y
y³ + 27/y³ - 12 = 0
moreover, substitute ; z = y³
z + 27/z - 12 = 0, z ≠ 0
z² - 12z + 27 = 0 ; z = 3, 9
either z = 3 or 9, we get same value
x = ∛3 + ∛9
the other two solutions are complex,
2y = (-1 + i√3)*∛z and (-1 - i√3)*∛z
Ooh... Wait. Did you just want to answer this question or also want to crack the equation?🤔😅
😮😄
if you actually try to do the second method the systems of equations require you to solve the original cubic so the second method is useless, why even mention it
What is the second method?
We do already know that you remember how to use Cardano method. No new ideas at all. Wasted time.
Thanks for the view 😁
@SyberMath but why start with a plus b to being with..ypu have to agree it comes out of nowhere...unless you first rule out integers and rational as solutions. .sona plus b represents a rational plus irrational number..is that why orif not why..I hope you can PLEASE PLEASE ANSWER
@vladimirkaplun5774: See my main comment on this video for an entirely different method to solve cubics which has nothing to do with the method named after Cardano and which you have probably never seen before.
@@NadiehFan Fresh view indeed.