How do you solve $8 x^{3} - 32 x = 0$ $8x^3 - 32x=0$ ?

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How do you solve $8 x^{3} - 32 x = 0$ $8x^3 - 32x=0$ ?

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Answer 1 · 2016-03-12T12:33:17

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

Explanation:

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

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How do you solve $8 x^{3} - 32 x = 0$ $8x^3 - 32x=0$ ?

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How do you solve 8x^3 - 32x=08x3−32x=08x^3 - 32x=0?

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How do you solve $8 x^{3} - 32 x = 0$ $8x^3 - 32x=0$ ?