What are the cube roots of 8?

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Answer 1 · 2018-03-04T16:12:21

Solve $x^3-8 = 0$

Use either known factoring $a^3-b^3 = (a-b)(a^2+ab+b^2)$ or divide $x^3-8$ by $x-2$ to find the quedratic factor.

$x^3-8 = (x-2)(x^2+2x+4)$

$x^3-8 = 0$ if and only if

$x-2 = 0$ or $x^2+2x+4 = 0$

So one solution is $2$

The quadratic may be solved by completing the square of by formula

$x = (-(2)+-sqrt((2)^2-4(1)(4)))/(2(1))$

$= -1+-isqrt3$