# How do you solve 8x^3 - 32x=0?

Mar 12, 2016

There is no need to use 'Comparing method' as suggested.
$x = 0 , - 2 , 2$

#### Explanation:

Given
$8 {x}^{3} - 32 x = 0$
Inspection reveals that $8 x$ is a factor of right hand side cubic.
$\therefore 8 x \left({x}^{2} - 4\right) = 0$
The quadratic term can be rewritten using the formula, since $4 = {2}^{2}$
${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
Hence the factors are
$8 x \left(x + 2\right) \left(x - 2\right) = 0$
Now $8 \ne 0$, setting remaining factors equal to zero, we obtain
$x = 0 , \left(x + 2\right) = 0 \mathmr{and} \left(x - 2\right) = 0$
We obtain three values of $x = 0 , - 2 , 2$