Help on Maths Question ?!!!!!!!!!!!!!!!!!!

Solve:
8x^3 + x^-3 = -9

Thanks!!!

8x^3 + x^-3 = - 9

First, add 9 to each side get an equation equal to 0:

8x^3 + x^-3 + 9 = 0

Now, multiply through by x^3 to eliminate the negative exponent:

8x^6 + 1 + 9x^3 = 0

Re-arrange:

8x^6 + 9x^3 + 1 = 0

Let u = x^3 and substitute:

8u^2 + 9u + 1 = 0

Factor by grouping:

8u^2 + 8u + u + 1 = 0

8u(u + 1) + 1(u + 1) = 0

(8u + 1) * (u + 1) = 0

Plug in u = x^3 again to get:

(8x^3 + 1) * (x^3 + 1) = 0

Now we have two solutions:

8x^3 + 1 = 0 and x^3 + 1 = 0

Solving each:

8x^3 = -1 and x^3 = -1

x^3 = -1/8 and x^3 = -1

By the cube roots, there are two solutions:

x = -1/2 and x = -1

8x^3 + x^-3 = -9
8x^3 + (1 / x^3) = -9
(8x^6 / x^3) + (1 / x^3) = -9
(8x^6 + 1) / x^3 = -9
8x^6 + 1 = -9x^3
8x^6 + 9x^3 + 1 = 0
(8x^3 + 1)(x^3 + 1) = 0
(2x + 1)(4x^2 - 2x + 1)(x + 1)(x^2 - x + 1) = 0
(2x + 1)(4x^2 - 2x + 1)(x + 1)(x^2 - x + 1) = 0

2x + 1 = 0
2x = -1
x = -1 / 2

x + 1 = 0
x = -1

4x^2 - 2x + 1 = 0
x = (-b +/- sqrt(b^2 - 4ac)) / 2a
x = (-(-2) +/- sqrt((-2)^2 - 4(4)(1))) / 2(4)
x = (2 +/- sqrt(4 - 16)) / 8
x = (2 +/- sqrt(-12)) / 8
x = (2 +/- sqrt(2^2 * 3 * -1)) / 8
x = (2 +/- 2i * sqrt(3)) / 8
x = (1 +/- i * sqrt(3)) / 4


x^2 - x + 1 = 0
x = (-b +/- sqrt(b^2 - 4ac)) / 2a
x = (-(-1) +/- sqrt((-1)^2 - 4(1)(1))) / 2(1)
x = (1 +/- sqrt(1 - 4)) / 2
x = (1 +/- sqrt(-3)) / 2
x = (1 +/- i * sqrt(3)) / 2



x = [-1, -1 / 2, (1 - i * sqrt(3)) / 4, (1 - i * sqrt(3)) / 2, (1 + i * sqrt(3)) / 4, (1 + i * sqrt(3)) / 2]

Math. Not MathS. That is all.