# How do you simplify (w+2t)(w^2-2wt+4t^2)?

##### 1 Answer
Jul 28, 2018

$\left(w + 2 t\right) \left({w}^{2} - 2 w t + 4 {t}^{2}\right) = {w}^{3} + 8 {t}^{3}$

#### Explanation:

The given expression is in the form:

$\left(A + B\right) \left({A}^{2} - A B + {B}^{2}\right)$

with $A = w$ and $B = 2 t$.

This is recognisable as the factored form (using real coefficients) of the sum of cubes:

${A}^{3} + {B}^{3} = \left(A + B\right) \left({A}^{2} - A B + {B}^{2}\right)$

If you wanted to multiply it out by hand, you could use distributivity as follows:

$\left(w + 2 t\right) \left({w}^{2} - 2 w t + 4 {t}^{2}\right) = w \left({w}^{2} - 2 w t + 4 {t}^{2}\right) + 2 t \left({w}^{2} - 2 w t + 4 {t}^{2}\right)$

$\textcolor{w h i t e}{\left(w + 2 t\right) \left({w}^{2} - 2 w t + 4 {t}^{2}\right)} = {w}^{3} - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 {w}^{2} t}}} + \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{4 w {t}^{2}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 {w}^{2} t}}} - \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{4 w {t}^{2}}}} + 8 {t}^{3}$

$\textcolor{w h i t e}{\left(w + 2 t\right) \left({w}^{2} - 2 w t + 4 {t}^{2}\right)} = {w}^{3} + 8 {t}^{3}$