A Nice Maths olympiad question | Algebra problem | (x+3)^6 = 2^6 |

[@berndfriedrichs6857]

using complex numbers, the 6 values for x can be found in seconds: x = -3 + 2*exp(i*2*pi*k/6) where k=0,1,2,3,4,5

[@Sweett88]

(x+3)^6 = 2^6
Square both sides by 1/6
x+3=2
Minus both sides by -3 and you got x=-1

[@rashel1]

thanks 😊

[@trevorsamaroo3623]

You need 6 solutions

[@reynaokeofficial_ph]

@@Sweett88 X can also actually be -5. (-5+3)^6 = (-2)^6 = switch to positive.

[@stpat7614]

I think my way is a bit easier:
(x + 3)^6 = 2^6
(x + 3)^(3 * 2) = 2^(3 * 2)
([x + 3]^3)^2 = (2^3)^2
Let a = (x + 3)^3, and b = 2^3
([x + 3]^3)^2 = (2^3)^2
=> a^2 = b^2
=> a^2 - b^2 = b^2 - b^2
=> a^2 - b^2 = 0
=> (a - b)(a + b) = 0
=> ([x + 3]^3 - 2^3)([x + 3]^3 + 2^3) = 0
Let a = x + 3, and b = 2
([x + 3]^3 - 2^3)([x + 3]^3 + 2^3) = 0
=> (a^3 - b^3)(a^3 + b^3) = 0
=> (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2) = 0
=> (a - b)(1 * a^2 + b * a + b^2)(a + b)(1 * a^2 + [-b] * a + b^2) = 0
Suppose a - b = 0
a - b = 0
a - b + b = 0 + b
a = b
Remember, a = x + 3, and b = 2
x + 3 = 2
x + 3 - 3 = 2 - 3
x = -1
x1 = -1
Suppose 1 * a^2 + b * a + b^2 = 0
1 * a^2 + b * a + b^2 = 0
a = (-b +/- sqrt[b^2 - 4 * 1 * b^2]) / (2 * 1)
a = (-b +/- sqrt[1 * b^2 - 4 * b^2]) / (2)
a = (-b +/- sqrt[(1 - 4) * b^2]) / 2
a = (-b +/- sqrt[(-3) * b^2]) / 2
a = (-b +/- sqrt[(-1) * 3 * b^2]) / 2
a = (-b +/- sqrt[-1] * sqrt[3] * sqrt[b^2]) / 2
a = (-b +/- i * sqrt[3] * b) / 2
a = (-1 +/- i * sqrt[3] * 1) * b / 2
Remember, a = x + 3, and b = 2
x + 3 = (-1 +/- i * sqrt[3]) * 2 / 2
x + 3 = (-1 +/- i * sqrt[3]) * 1
x + 3 = -1 +/- i * sqrt(3)
x + 3 - 3 = -3 - 1 +/- i * sqrt(3)
x = -4 +/- i * sqrt(3)
x = -4 + i * sqrt(3), or x = -4 - i * sqrt(3)
x2 = -4 + i * sqrt(3)
x3 = -4 - i * sqrt(3)
Suppose a + b = 0
a + b = 0
a + b - b = 0 - b
a = -b
Remember, a = x + 3, and b = 2
x + 3 = -2
x + 3 - 3 = -2 - 3
x = -5
x4 = -5
Suppose 1 * a^2 + (-b) * a + b^2 = 0
1 * a^2 + (-b) * a + b^2 = 0
a = (-[-b] +/- sqrt[(-b)^2 - 4 * 1 * b^2]) / (2 * 1)
a = (b +/- sqrt[1 * b^2 - 4 * b^2]) / (2)
a = (b +/- sqrt[(1 - 4) * b^2]) / 2
a = (b +/- sqrt[(-3) * b^2]) / 2
a = (b +/- sqrt[(-1) * 3 * b^2]) / 2
a = (b +/- sqrt[-1] * sqrt[3] * sqrt[b^2]) / 2
a = (b +/- i * sqrt[3] * b) / 2
a = (1 +/- i * sqrt[3] * 1) * b / 2
Remember, a = x + 3, and b = 2
x + 3 = (1 +/- i * sqrt[3]) * 2 / 2
x + 3 = (1 +/- i * sqrt[3]) * 1
x + 3 = 1 +/- i * sqrt(3)
x + 3 = 1 + i * sqrt(3), or x + 3 = 1 - i * sqrt(3)
x + 3 - 3 = -3 + 1 + i * sqrt(3), or x + 3 - 3 = -3 + 1 + i * sqrt(3)
x = -2 - i * sqrt(3), or x = -2 + i * sqrt(3)
x5 = -2 - i * sqrt(3)
x6 = -2 + i * sqrt(3)
x1 = -1
x2 = -4 + i * sqrt(3)
x3 = -4 - i * sqrt(3)
x4 = -5
x5 = -2 - i * sqrt(3)
x6 = -2 + i * sqrt(3)

[@WasifRBX]

I don't know if my procedure is correct but:
The powers are same so cancel power of 6
x + 3 = 2
x = 2 - 3
x = -1

[@davidseed2939]

no you need to use 6 roots of unity
z= 1,-1, +-1/2 +-isqrt(3)/2
x=-3+2z
draw on complex plane so see whats happening

[@berndfriedrichs6857]

Many comments seem to ignore the basic fact that every polynomial of degree n has exactly n complex roots. In this case, there exist two real-valued roots (-1 and -5) and the other four roots are complex-valued (two pairs of complex conjugated roots, see my other comment below)

[@achyuththouta6957]

They are suffering from the dunning Kruger effect

[@88kgs]

Thank you for sharing and explaining in detail.
Enjoying school days after 30 years..
Regards 🙏

[@ranajoybardhan7197]

You can do it like
(X+3)^6=2^6
6 will be subtract from both sides
X+3=2
X=-1

[@phanisharma9978]

It's so simple.

[@rashel1]

I'm glad you found it easy!

[@reynaokeofficial_ph]

X = -1 (simple)

[@chezzaization]

I got x=1

[@MuhammadSulemanSulu]

It was a very simple question you can say only one sentence question and answer was you know hidden in the question basically but you just stretch the question to lengthy and I am supposed to be have the headache you know by watching you what was the thing that make you so much you know in difference to this

[@rashel1]

Thanks for the feedback! I try to explain things in a way that's easy to understand. 😊

[@ju2705s]

3s X=-1

[@Syife-n2z]

X= -1

[@yalcinhamdioglu5916]

X=-5 or x=-1

[@joshuabjk9321]

Tf such a useless