How do you factor: y= 32x^3 - 4 ?

2 Answers
Apr 14, 2017

$y = 4 \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$

Explanation:

Recall:

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

Here we go.

$y = 32 {x}^{3} - 4$

By factoring out $4$,

$R i g h t a r r o w y = 4 \left(8 {x}^{3} - 1\right)$

By rewriting a bit,

$R i g h t a r r o w y = 4 \left({\left(2 x\right)}^{3} - {1}^{3}\right)$

By applying the formula at the top with $a = 2 x$ and $b = 1$,

$R i g h t a r r o w y = 4 \left(2 x - 1\right) \left({\left(2 x\right)}^{2} + 2 x \cdot 1 + {1}^{2}\right)$

By cleaning up a bit,

$y = 4 \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$

I hope that this was clear.

Apr 14, 2017

$y = 4 \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$

Explanation:

The first step is to factor out $\textcolor{b l u e}{\text{common factor}}$ of 4

$\Rightarrow y = 4 \left(8 {x}^{3} - 1\right) \to \left(1\right)$

now $8 {x}^{3} - 1$ is a $\textcolor{b l u e}{\text{difference of cubes}}$ and factorises in general as.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$8 {x}^{3} = {\left(2 x\right)}^{3} \text{ and } {1}^{3} = 1$

$\text{using " a=2x" and } b = 1$

$\Rightarrow 8 {x}^{3} - 1 = \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$

$\text{going back to } \left(1\right)$

$\Rightarrow y = 4 \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$