# How do you find the real or imaginary solutions of the equation x^3+64=0?

Nov 27, 2016

Use the sum of cubes identity to find the Real zero and a quadratic to solve and find the zeros are:

$4 \text{ }$ and $\text{ } 2 \pm 2 \sqrt{3} i$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this later with $a = \left(x - 2\right)$ and $b = \sqrt{- 12} = 2 \sqrt{3} i$, but first...

The sum of cubes identity can be written:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Note that ${x}^{3}$ and $64 = {4}^{3}$ are both perfect squares, so the sub of cubes identity applies directly:

${x}^{3} + 64 = {x}^{3} + {4}^{3}$

$\textcolor{w h i t e}{{x}^{3} + 64} = \left(x + 4\right) \left({x}^{2} - 4 x + 16\right)$

We can factor the quadratic by completing the square:

${x}^{2} - 4 x + 16 = {x}^{2} - 4 x + 4 + 12$

$\textcolor{w h i t e}{{x}^{2} - 4 x + 16} = {\left(x - 2\right)}^{2} - {\left(2 \sqrt{3} i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 4 x + 16} = \left(\left(x - 2\right) - 2 \sqrt{3} i\right) \left(\left(x - 2\right) + 2 \sqrt{3} i\right)$

$\textcolor{w h i t e}{{x}^{2} - 4 x + 16} = \left(x - 2 - 2 \sqrt{3} i\right) \left(x - 2 + 2 \sqrt{3} i\right)$

Hence the zeros of ${x}^{3} + 64$ are:

$4 \text{ }$ and $\text{ } 2 \pm 2 \sqrt{3} i$