more complicated than it need have been. m had to be positive to allow real sqrt(m). in which case y must be positive so case 2 is the only possible solution. Also. evaluating m from y. we did not need to solve any equation we just needed arithmetic to find (( sqrt3-1)/3)^6
Substitution m=y^6 is a not equivalent transformation and could produce solution outside the domain of m, m^1/2, and m^1/3. The domain of y^6 should be established. From m^1/2 >=0 => (y^6)^1/2 >= 0 => y^3>=0 => y>=0. (checking the domain of m^1/3 and the domain of m will not provide any further restrictions to the domain of y). Since y is positive the negative solutions y1 and y3 are rejected. The domains for substitution m=y^(1/6) proposed in comments shall be checked in the same way.
sale un poco más rápido y más intuitivo completando el cubo, queda una expresión donde se ve claramente la factorización: (y + 1/3)^3 - 1/3 (y + 1/3) = 0
Substitution: y=m^(1/6). Then the original equation implies y^3+y^2-2/27=0. [1] [1] has an obvious root y=-1/3. Therefore [1] factors as (y+1/3)(y+2y/3-2/9)=0 [2] by synthetic divison. [2] has three roots: -1/3 , (-1+√3)/3, (-1-√3)/3. Therefore m=1/729 or m=[(-1+√3)/3]^6 or m=[(-1-√3)/3]**6. By inspection only m=[(-1+√3)/3]^6 solves ∛m+√m=2/27.
I did it exactly the same. I didn't see the rational root and substituted n^6=m then u=n-1/3 which led me to u belongs to {0,√3/3,-√3/3} then we get n and m. (1/3)^6 is not a root because 1/9+1/27 2/27 ((√3+1)/3)^2 + ((√3+1)/3)^3 obviously belongs to R[√3] as there is no way for √3 to cancel not sure why video is so long
@@mab9316 Yes, there is: [1] is implied by the original equation if y=m^(1/6), but [1[ and y=m^(1/6) do not imply the original equation. Thus, for all solutions m of the original equation, y=m^(1/6) will be a solution of [1], but not for all solutions y of [1] m=y^6 will necessarily be a solution of the original equation. This is similar to x+√x-1=0 and y=x^(1/2) ==> y^2+y-1=0. There are two solutions y1=(-1-√5)/2 and y2=(-1+√5)/2, but only x=y2^2 is a solution for the first equation.
more complicated than it need have been. m had to be positive to allow real sqrt(m).
in which case y must be positive so case 2 is the only possible solution.
Also. evaluating m from y. we did not need to solve any equation we just needed arithmetic to find (( sqrt3-1)/3)^6
GREAT WORKS IN YOUR MASTERFUL CALCULATION, especially in the very last segment of the m[value]...
Soln:
m^1/3 + m^1/2 = 2/27
(m^1/3) + (m^1/3)^3/2 = 2/27
Let u = m^1/3
u + u^3/2 = 2/27
(u^1/2)^2 + (u^1/2)^3 = 2/27. Let y = u^1/2
y^2 + y^3 =2/27
y^3 + y^2 - 2/27 = 0 [x 27 both sides]
27y^3 + 27y^2 - 2 = 0
27y^3 + 18y^2 - 6y + 9y^2 + 6y - 2 = 0
3y(9y^2+6y-2)+1(9y^2+6y-2)=0
(3y+1)(9y^2+6y-2) = 0
3y+1 =0. 9y^2+6y-2=0
3y = -1
y = -1/3
2nd[eqn]
y = -6 +/- sqr 108/18
= -6 +/- sqr36.sqr3/18
= -6+/- 6sqr3/18
= -1 +/- sqr3/3
y = sqr3 - 1/3. Or
-sqr3 - 1/3
u^1/2 = -1/3
u = (-1/3)^2
= 1/9
u^1/2 = sqr3 - 1/3
u = [ sqr3 - 1/3]^2
= 3 - 2sqr3 + 1/9
u = 4 - 2sqr3/9
u^1/2 = -sqr3 - 1/3
u = [-sqr3 - 1/3]^2
= 3 + 2sqr3 + 1/9
u = 4 +2sqr3/9
u = m^1/3. ; u = 1/9
m^1/3 = 1/9
m = (1/9)^3
m = 1/729
Check: 1/9 + 1/27
= 3 + 1/27 = 2/27
4/27 = 2/27
Therefore m=1/729 is not valid
u = 4 - 2sqr3/9
m^1/3 = u
m^1/3 = (4 - 2sqr3/9)^3
m = 16 - 8sqr3 - 8sqr3 + 12[4 - 2sqr3] /729
=16-32sqr3+12[4 - 2sqr3]/729
= 28 -16sqr3[4 -2sqr3]/729
= 112 - 56sqr3 - 64sqr3 + 96/729
= 112+96-120sqr3/729
= 208 - 120sqr3 /729
Check: m^1/3 + m^1/2 = 2/27
0.0595 + 0.0145 = 2/27
= 0.074 = 2/27 = 0.0740
Therefore :
208 - 120 sqr 3 / 729
is valid
m^1/3 = 4 + 2sqr3 /9
m = [ 4 + 2sqr3/9 ] ^3
m = 208 + 120sqr3/729
This result is invalid....
Substitution m=y^6 is a not equivalent transformation and could produce solution outside the domain of m, m^1/2, and m^1/3. The domain of y^6 should be established.
From m^1/2 >=0 => (y^6)^1/2 >= 0 => y^3>=0 => y>=0. (checking the domain of m^1/3 and the domain of m will not provide any further restrictions to the domain of y).
Since y is positive the negative solutions y1 and y3 are rejected.
The domains for substitution m=y^(1/6) proposed in comments shall be checked in the same way.
m^1/3 + m^1/2 = 2/27
m^2/6 + m^3/6 = 2/27
{m^1/6}^2 + {m^1/6}^3 = 1/9 - 1/27
{m^1/6}^2 + {m^1/6}^3 = (1/3)^2 - (1/3)^3
let m^1/6 be a.
a^2 + a^3 - (1/3)^2 + (1/3)^3 = 0
use a^2 - b^2 and a^3 + b^3 formula and factorize.
sale un poco más rápido y más intuitivo completando el cubo, queda una expresión donde se ve claramente la factorización: (y + 1/3)^3 - 1/3 (y + 1/3) = 0
Beautiful
Thanks 💯😎👏🙏
Substitution: y=m^(1/6). Then the original equation implies
y^3+y^2-2/27=0. [1]
[1] has an obvious root y=-1/3. Therefore [1] factors as
(y+1/3)(y+2y/3-2/9)=0 [2] by synthetic divison.
[2] has three roots: -1/3 , (-1+√3)/3, (-1-√3)/3.
Therefore m=1/729 or m=[(-1+√3)/3]^6 or m=[(-1-√3)/3]**6.
By inspection only m=[(-1+√3)/3]^6 solves ∛m+√m=2/27.
Is there any explanation why some roots being rejected, even though they are solutions to the equation ???
I did it exactly the same. I didn't see the rational root and substituted
n^6=m then u=n-1/3 which led me to u belongs to {0,√3/3,-√3/3} then we get n and m.
(1/3)^6 is not a root because 1/9+1/27 2/27
((√3+1)/3)^2 + ((√3+1)/3)^3 obviously belongs to R[√3] as there is no way for √3 to cancel
not sure why video is so long
@@mab9316 we substituted with a square that changes sign.
But in original equation we either lose the sign or we get complex or extraneous solutions.
@@mab9316 Yes, there is:
[1] is implied by the original equation if y=m^(1/6), but [1[ and y=m^(1/6) do not imply the original equation.
Thus, for all solutions m of the original equation, y=m^(1/6) will be a solution of [1], but not for all solutions y of [1] m=y^6 will necessarily be a solution of the original equation.
This is similar to x+√x-1=0 and y=x^(1/2) ==> y^2+y-1=0. There are two solutions y1=(-1-√5)/2 and y2=(-1+√5)/2,
but only x=y2^2 is a solution for the first equation.
@@mab9316 Please see my comment
M= (-1/3)^6 is
also a solution
At start it looks complex but after u did the first 3 steps it’s soo easy. Tysm ur soo great. I❤algebra and u made me fall for it more
Great job! Happy to help. 😁🙏🤩
It was well explained--'thinking outside of the box'.
Thanks ✅✅✅
m^1/3 + m^1/2 = 2/27
(m^1/3) + (m^1/3)^3/2 = 2/27
Let u = m^1/3
u + u^3/2 = 2/27
(u^1/2)^2 + (u^1/2)^3 = 2/27. Let y = u^1/2
y^2 + y^3 =2/27
y^3 + y^2 - 2/27 = 0 [x 27 both sides]
27y^3 + 27y^2 - 2 = 0
27y^3 + 18y^2 - 6y + 9y^2 + 6y - 2 = 0
3y(9y^2+6y-2)+1(9y^2+6y-2)=0
(3y+1)(9y^2+6y-2) = 0
3y+1 =0. 9y^2+6y-2=0
3y = -1
y = -1/3
2nd[eqn]
y = -6 +/- sqr 108/18
= -6 +/- sqr36.sqr3/18
= -6+/- 6sqr3/18
= -1 +/- sqr3/3
y = sqr3 - 1/3. Or
-sqr3 - 1/3
u^1/2 = -1/3
u = (-1/3)^2
= 1/9
u^1/2 = sqr3 - 1/3
u = [ sqr3 - 1/3]^2
= 3 - 2sqr3 + 1/9
u = 4 - 2sqr3/9
u^1/2 = -sqr3 - 1/3
u = [-sqr3 - 1/3]^2
= 3 + 2sqr3 + 1/9
u = 4 +2sqr3/9
u = m^1/3. ; u = 1/9
m^1/3 = 1/9
m = (1/9)^3
m = 1/729
Check: 1/9 + 1/27
= 3 + 1/27 = 2/27
4/27 = 2/27
Therefore m=1/729 is not valid
u = 4 - 2sqr3/9
m^1/3 = u
m^1/3 = (4 - 2sqr3/9)^3
m = 16 - 8sqr3 - 8sqr3 + 12[4 - 2sqr3] /729
=16-32sqr3+12[4 - 2sqr3]/729
= 28 -16sqr3[4 -2sqr3]/729
= 112 - 56sqr3 - 64sqr3 + 96/729
= 112+96-120sqr3/729
= 208 - 120sqr3 /729
Check: m^1/3 + m^1/2 = 2/27
0.0595 + 0.0145 = 2/27
= 0.074 = 2/27 = 0.0740
Therefore :
208 - 120 sqr 3 / 729
is valid
solution
Awesome 😎💯🙏✅
🎉❤😂 👍
Easy
³√m + √m = 2/27
³√(u⁶) + √(u⁶) = 2/27 --- m = u⁶
u² + u³ = 2/3³
u³ + u² - 2/3³ = 0
u³ + u² - 3/3³ + 1/3³ = 0
u³ + u² - 1/3² + 1/3³ = 0
(u³+1/3³) + (u²-1/3²) = 0
(u+1/3)(u²-u/3+1/9) + (u+1/3)(u-1/3) = 0
(u+1/3)(u²-u/3+1/9+u-1/3) = 0
(u+1/3)(u²+2u/3-2/9) = 0
u = -1/3 ✓ [¹] | u² + 2u/3 - 2/9 = 0
9u² + 6u - 2 = 0
u = [-(6)±√((6)²-4(9)(-2))]/2(9)
u = -6/18 ± √(36+72)/18
u = -1/3 ± √108/18
u = -1/3 ± 6√3/18 = (-1±√3)/3
u = -(1-√3)/3 ✓ [²] | u = -(1+√3)/3 ✓ [³]
[1]
m = (-1/3)⁶ = 1/(-3)⁶ = 1/9³ = 1/729
³√(1/729) + √(1/729) = 2/27
1/9 + 1/27 = 2/27
4/27 ≠ 2/27 ❌ m = 1/729 invalid
[2]
m = (-(1-√3)/3)⁶ = (√3-1)⁶/3⁶
³√((√3-1)⁶/3⁶) + √((√3-1)⁶/3⁶) = 2/27
(√3-1)²/3² + (√3-1)³/3³ = 2/27
(3-2√3+1)/9 + (3√3-9+3√3-1)/27 = 2/27
(12-6√3)/27 + (6√3-10)/27 = 2/27
(12-10)/27 = 2/27
2/27 = 2/27 ✓ m = (-(1-√3)/3)⁶ valid
m = ((6√3-10)/27)²
m = (108-120√3+100)/729
m = (208-120√3)/729
m = 8(26-15√3)/729 ≈ 2.111×10⁻⁴
[3]
m = (-(1+√3)/3)⁶ = (√3+1)⁶/(-3)⁶
³√((√3+1)⁶/(-3)⁶) + √((√3+1)/(-3)⁶) = 2/27
(√3+1)²/3² + (√3+1)³/3³ = 2/27
(3+2√3+1)/9 + (3√3+9+3√3+1) = 2/27
(12+6√3)/27 + (6√3+10)/27 = 2/27
(22+12√3)/27 ≠ 2/27 ❌
m = (-(1+√3)/3)⁶ invalid
m = 8(26-15√3)/729 ≈ 2.111×10⁻⁴
There are 2 square roots of 729. -27 provides the correct answer.
{m+m+m+m ➖ m ➖ m} +{ m+m+m m ➖ m ➖ m}= {m^3+m^3}=m^6 m^3^3(m ➖ 3m+3).
m^3+m^3 = 1m^3 and not m^6!!!!!!!!!!!!!!!!!!!!!!!!