Germany | Can you solve this? | Math Olympiad

[@stpat7614]

Slightly different (and easier) method:
(x - 2)^6 = 5^6
(x - 2)^(3*2) = 5^(3*2)
([x - 2]^3)^2 = (5^3)^2
Let a = (x - 2)^3, and b = 5^3
([x - 2]^3)^2 = (5^3)^2
=> a^2 = b^2
=> a^2 - b^2 = b^2 - b^2
=> a^2 - b^2 = 0
=> (a - b)(a + b) = 0
=> ([x - 2]^3 - 5^3)([x - 2]^3 + 5^3) = 0
Let a = x - 2, and b = 5
([x - 2]^3 - 5^3)([x - 2]^3 + 5^3) = 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
Remember, a = x - 2, and b = 5
(x - 2) - 5 = 0
x - 2 - 5 = 0
x - 7 = 0
x - 7 + 7 = 0 + 7
x = 7
x1 = 7
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[b^2 * (1 - 4)]) / (2)
a = (-b +/- sqrt[b^2 * (-3)]) / 2
a = (-b +/- sqrt[b^2 * 3 * (-1)]) / 2
a = (-b +/- sqrt[b^2] * sqrt[3] * sqrt[-1]) / 2
a = (-b +/- b * sqrt[3] * i) / 2
Remember, a = x - 2, and b = 5
x - 2 = (-5 +/- 5 * sqrt[3] * i) / 2
x - 2 + 2 = 2 + (-5 +/- 5 * sqrt[3] * i) / 2
x = 2 + (-5 +/- 5 * sqrt[3] * i) / 2
x = 2 * 2 / 2 + (-5 +/- 5 * sqrt[3] * i) / 2
x = 4 / 2 + (-5 +/- 5 * sqrt[3] * i) / 2
x = (4 - 5 +/- 5 * sqrt[3] * i) / 2
x = ([4 - 5] +/- 5 * sqrt[3] * i) / 2
x = (-1 +/- 5 * sqrt[3] * i) / 2
x = (-1 + 5 * sqrt[3] * i) / 2, or x = (-1 - 5 * sqrt[3] * i) / 2
x = -(1 - 5 * sqrt[3] * i) / 2, or x = -(1 + 5 * sqrt[3] * i) / 2
x2 = -(1 - 5 * sqrt[3] * i) / 2
x3 = -(1 + 5 * sqrt[3] * i) / 2
Suppose a + b = 0
a + b = 0
Remember, a = x - 2, and b = 5
(x - 2) + 5 = 0
x - 2 + 5 = 0
x + 3 = 0
x + 3 - 3 = 0 - 3
x = -3
x4 = -3
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)^2 * b^2 - 4 * b^2]) / (2)
a = (b +/- sqrt[1 * b^2 - 4 * b^2]) / 2
a = (b +/- sqrt[b^2 * 1 - b^2 * 4]) / 2
a = (b +/- sqrt[b^2 * (1 - 4)]) / 2
a = (b +/- sqrt[b^2 * (-3)]) / 2
a = (b +/- sqrt[b^2 * 3 * (-1)]) / 2
a = (b +/- sqrt[b^2] * sqrt[3] * sqrt[-1]) / 2
a = (b +/- b * sqrt[3] * i) / 2
Remember, a = x - 2, and b = 5
x - 2 = (5 +/- 5 * sqrt[3] * i) / 2
x - 2 + 2 = 2 + (5 +/- 5 * sqrt[3] * i) / 2
x = 2 + (5 +/- 5 * sqrt[3] * i) / 2
x = 2 * 2 / 2 + (5 +/- 5 * sqrt[3] * i) / 2
x = 4 / 2 + (5 +/- 5 * sqrt[3] * i) / 2
x = (4 + 5 +/- sqrt[3] * i) / 2
x = ([4 + 5] +/- 5 * sqrt[3] * i) / 2
x = (9 +/- 5 * sqrt[3] * i) / 2
x = (9 + 5 * sqrt[3] * i) / 2, or x = (9 - 5 * sqrt[3] * i) / 2
x5 = (9 + 5 * sqrt[3] * i) / 2
x6 = (9 - 5 * sqrt[3] * i) / 2
{x1, x2, x3, x4, x5, x6} = {7, -(1 - 5 * sqrt[3] * i) / 2, -(1 + 5 * sqrt[3] * i) / 2, -3, (9 + 5 * sqrt[3] * i) / 2, (9 - 5 * sqrt[3] * i) / 2}

[@bonilsson6349]

JI don't understand anything... Remove the exponent in both joints so you have X-2=5 => X = 7 ????? ...

[@Syife-n2z]

كما فعلت أنا

[@santer70]

@bonilsson6349 the problem is that you have to find 6 solution for X cause the exponent is ^6

[@waltertonelli2174]

you must add x-2=-5 -> x=-3 but only if you are looking for real solutions

[@JagoKhan-t4o]

X=7 is correct 😡😡