# How do you solve x^2(a^2+2ab+b^2) = x(a+b) by factoring?

Aug 21, 2015

$x = 0 \text{ }$ or $\text{ } x = \frac{1}{a + b}$

#### Explanation:

First, notice that you can write

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

as the square of $\left(a + b\right)$

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

Your equation can now be written as

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

You can simplify this equation by dividing both sides by $\left(a + b\right)$

$\frac{{x}^{2} \cdot {\left(a + b\right)}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(a + b\right)}}}} = \frac{x \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(a + b\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(a + b\right)}}}}$

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

Move all the terms to one side of the equation to get

$\left(a + b\right) \cdot {x}^{2} - x = 0$

$x \cdot \left[\left(a + b\right) x - 1\right] = 0$

This equation will have two solutions, $x = \textcolor{g r e e n}{0}$ and

$\left(a + b\right) x - 1 = 0 \implies x = \textcolor{g r e e n}{\frac{1}{a + b}}$