You need to solve for x the equation`a^3x^3 + b^3 = 0` , hence, you need to convert the sum of cubes into a product such that:

`a^3x^3 + b^3 = (ax + b)(a^2x^2- abx + b^2)`

`a^3x^3 + b^3 = 0 => (ax + b)(a^2x^2 - abx + b^2)...

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You need to solve for x the equation`a^3x^3 + b^3 = 0` , hence, you need to convert the sum of cubes into a product such that:

`a^3x^3 + b^3 = (ax + b)(a^2x^2- abx + b^2)`

`a^3x^3 + b^3 = 0 => (ax + b)(a^2x^2 - abx + b^2) = 0`

You need to set the factors of product equal to zero such that:

`ax + b = 0 => ax = -b => x = -b/a`

`a^2x^2 - abx + b^2 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (ab+-sqrt(a^2b^2 - 4a^2b^2))/(2a^2)`

`x_(1,2) = (ab+-sqrt(-3a^2b^2))/(2a^2)`

Since `sqrt -1 = i ` (complex number theory), yields:

`x_(1,2) = (ab+-ab*sqrt3*i)/(2a^2)`

Factoring out a yields:

`x_(1,2) = (a(b+-b*sqrt3*i))/(2a^2) => x_(1,2) = (b+-b*sqrt3*i)/(2a) `

**Hence, evaluating the roots of the equation `a^3x^3 + b^3 = 0` yields one real root `x = -b/a` and two complex conjugate roots `x_(1,2) = (b+-b*sqrt3*i)/(2a).` **