# How do you simplify (x^2-6x+9)/(81-x^4)?

Jun 6, 2016

$\frac{{x}^{2} - 6 x + 9}{81 - {x}^{4}} = \frac{3 - x}{\left(9 + {x}^{2}\right) \left(3 + x\right)}$

#### Explanation:

To simplify $\frac{{x}^{2} - 6 x + 9}{81 - {x}^{4}}$, first factorize polynomials in numerator and denominator.

${x}^{2} - 6 x + 9$ is complete square of the type ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$, as ${x}^{2} - 6 x + 9 = {\left(x\right)}^{2} - 2 \times x \times 3 + {3}^{2}$. Hence,

${x}^{2} - 6 x + 9 = {\left(x - 3\right)}^{2}$.

This can also be written as $9 - 6 x + {x}^{2} = {\left(3 - x\right)}^{2}$

For factorizing $81 - {x}^{4}$, we use the identity $\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$

Hence 81-x^4=(9)^2-*x^2)^2=(9+x^2)(9+x^2)=(9+x^2)(3^2-x^2)

= $\left(9 + {x}^{2}\right) \left(3 + x\right) \left(3 - x\right)$

Hence $\frac{{x}^{2} - 6 x + 9}{81 - {x}^{4}} = {\left(3 - x\right)}^{2} / \left(\left(9 + {x}^{2}\right) \left(3 + x\right) \left(3 - x\right)\right)$

= $\frac{3 - x}{\left(9 + {x}^{2}\right) \left(3 + x\right)}$