How do you simplify (x-3)/(x-5) div (x^3-27)/(x^3+8)*(x^2-25)/(x + 5)?

Mar 15, 2016

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

Explanation:

Recall the following:

$1$. Difference of cubes: $\textcolor{\mathmr{and} a n \ge}{\left({a}^{3} - {b}^{3}\right) = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)}$

$2$. Sum of cubes: $\textcolor{b l u e}{\left({a}^{3} + {b}^{3}\right) = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)}$

$3$. Difference of squares: $\textcolor{p u r p \le}{\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)}$

The Simplification Process
$1$. Start by factoring the polynomials in the numerator and denominator of the second fraction.

$\frac{x - 3}{x - 5} \div \frac{{x}^{3} - 27}{{x}^{3} + 8} \cdot \frac{{x}^{2} - 25}{x + 5}$

$= \frac{x - 3}{x - 5} \div \frac{\textcolor{\mathmr{and} a n \ge}{\left(x - 3\right) \left({\left(x\right)}^{2} + \left(3\right) \left(x\right) + {\left(3\right)}^{2}\right)}}{\textcolor{b l u e}{\left(x + 2\right) \left({\left(x\right)}^{2} - \left(2\right) \left(x\right) + {\left(2\right)}^{2}\right)}} \cdot \frac{{x}^{2} - 25}{x + 5}$

$2$. Simplify.

$= \frac{x - 3}{x - 5} \div \frac{\textcolor{\mathmr{and} a n \ge}{\left(x - 3\right) \left({x}^{2} + 3 x + 9\right)}}{\textcolor{b l u e}{\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)}} \cdot \frac{{x}^{2} - 25}{x + 5}$

$3$. Factor the polynomials in the numerator and denominator of the third fraction.

$= \frac{x - 3}{x - 5} \div \frac{\textcolor{\mathmr{and} a n \ge}{\left(x - 3\right) \left({x}^{2} + 3 x + 9\right)}}{\textcolor{b l u e}{\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)}} \cdot \frac{\textcolor{p u r p \le}{\left(x + 5\right) \left(x - 5\right)}}{x + 5}$

$4$. To divide the first fraction by the second fraction, rewrite the second fraction as its reciprocal.

$= \frac{x - 3}{x - 5} \cdot \frac{\textcolor{b l u e}{\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)}}{\textcolor{\mathmr{and} a n \ge}{\left(x - 3\right) \left({x}^{2} + 3 x + 9\right)}} \cdot \frac{\textcolor{p u r p \le}{\left(x + 5\right) \left(x - 5\right)}}{x + 5}$

$5$. Cancel out any factors which appear in the numerator and denominator as a pair.

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 3\right)}}}}{\textcolor{t e a l}{\cancel{\textcolor{b l a c k}{\left(x - 5\right)}}}} \cdot \frac{\textcolor{b l u e}{\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)}}{\textcolor{\mathmr{and} a n \ge}{\textcolor{red}{\cancel{\textcolor{\mathmr{and} a n \ge}{\left(x - 3\right)}}} \left({x}^{2} + 3 x + 9\right)}} \cdot \frac{\textcolor{p u r p \le}{\textcolor{m a \ge n t a}{\cancel{\textcolor{p u r p \le}{\left(x + 5\right)}}} \textcolor{t e a l}{\cancel{\textcolor{p u r p \le}{\left(x - 5\right)}}}}}{\textcolor{m a \ge n t a}{\cancel{\textcolor{b l a c k}{\left(x + 5\right)}}}}$

$6$. Rewrite the expression.

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)}{{x}^{2} + 3 x + 9} \textcolor{w h i t e}{\frac{a}{a}} |}}}$