# How do you simplify (1 - x^2)^(1/2) - x^2(1 - x^2)^(-3/2)?

Jul 4, 2015

$\frac{\left(- {x}^{2} + x + 1\right) \left(- {x}^{2} - x + 1\right)}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

#### Explanation:

${\left(1 - {x}^{2}\right)}^{\frac{1}{2}} - {x}^{2} {\left(1 - {x}^{2}\right)}^{- \frac{3}{2}}$

We will use : $\textcolor{red}{{a}^{- n} = \frac{1}{a} ^ n}$

$\iff {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} - {x}^{2} / {\left(1 - {x}^{2}\right)}^{\textcolor{red}{+ \frac{3}{2}}}$

We want two fractions with the same denominator.

$\iff \frac{{\left(1 - {x}^{2}\right)}^{\frac{1}{2}} \cdot \textcolor{g r e e n}{{\left(1 - {x}^{2}\right)}^{\frac{3}{2}}}}{\textcolor{g r e e n}{{\left(1 - {x}^{2}\right)}^{\frac{3}{2}}}} - {x}^{2} / {\left(1 - {x}^{2}\right)}^{+ \frac{3}{2}}$

We will use : $\textcolor{red}{{u}^{a} \cdot {u}^{b} = {u}^{a + b}}$

$\iff \frac{\textcolor{red}{{\left(1 - {x}^{2}\right)}^{2}}}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right) - {x}^{2} / {\left(1 - {x}^{2}\right)}^{\frac{3}{2}}$

$\iff \frac{{\left(1 - {x}^{2}\right)}^{2} - {x}^{2}}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

We will use the following polynomial identity :

$\textcolor{b l u e}{\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}}$

$\iff \frac{\textcolor{b l u e}{\left(1 - {x}^{2} + x\right) \left(1 - {x}^{2} - x\right)}}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

$\iff \frac{\left(- {x}^{2} + x + 1\right) \left(- {x}^{2} - x + 1\right)}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

We can't do better than this, and now you can easily (if you want) find the solution of $\frac{\left(- {x}^{2} + x + 1\right) \left(- {x}^{2} - x + 1\right)}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right) = 0$