# How do you simplify [(x-a)+(2a^2)/(x+a)] * [((1/x^2)-(1/a^2))] div[((x^3)-((ax^3+a^4)/(x+a)))]?

Mar 30, 2017

$- \frac{1}{{x}^{2} {a}^{2}} .$

#### Explanation:

The Exp.$= \left[\frac{\left(x - a\right) \left(x + a\right) + 2 {a}^{2}}{x + a}\right] \cdot \left[\frac{{a}^{2} - {x}^{2}}{{x}^{2} {a}^{2}}\right] \div \left[{x}^{3} - \left\{\frac{a \left({x}^{3} + {a}^{3}\right)}{x + a}\right\}\right] ,$

$= \left[\frac{\left({x}^{2} - {a}^{2}\right) + 2 {a}^{2}}{x + a}\right] \cdot \left[\frac{\left(a + x\right) \left(a - x\right)}{{x}^{2} {a}^{2}}\right] \div \left[{x}^{3} - \frac{a \left(x + a\right) \left({x}^{2} - x a + {a}^{2}\right)}{x + a}\right] ,$

$= \left({x}^{2} + {a}^{2}\right) \cdot \left\{\frac{a - x}{{x}^{2} {a}^{2}}\right\} \div \left[{x}^{3} - a \left({x}^{2} - x a + {a}^{2}\right)\right] ,$

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

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

$= \frac{\left({x}^{2} + {a}^{2}\right) \left(a - x\right)}{{x}^{2} {a}^{2}} \div \left\{\left(x - a\right) \left({x}^{2} + x a + {a}^{2}\right) - x a \left(x - a\right)\right\} ,$

$= \frac{\left({x}^{2} + {a}^{2}\right) \left(a - x\right)}{{x}^{2} {a}^{2}} \div \left\{\left(x - a\right) \left({x}^{2} + x a + {a}^{2} - x a\right)\right\} ,$

$= \frac{\left({x}^{2} + {a}^{2}\right) \left(a - x\right)}{{x}^{2} {a}^{2}} \div \left\{\left(x - a\right) \left({x}^{2} + {a}^{2}\right)\right\} ,$

$= \frac{\left({x}^{2} + {a}^{2}\right) \left(a - x\right)}{{x}^{2} {a}^{2}} \times \frac{1}{- \left(a - x\right) \left({x}^{2} + {a}^{2}\right)} ,$

$\Rightarrow \text{ The Exp.=} - \frac{1}{{x}^{2} {a}^{2}} .$

Enjoy Maths.!

Mar 30, 2017

$- \frac{1}{{a}^{2} {x}^{2}}$

#### Explanation:

Calling

${f}_{1} = x - a + \frac{2 {a}^{2}}{x + a} = \frac{{x}^{2} - {a}^{2} + 2 {a}^{2}}{x + a} = \frac{{x}^{2} + {a}^{2}}{x + a}$

${f}_{2} = \frac{1}{x} ^ 2 - \frac{1}{a} ^ 2 = \frac{{a}^{2} - {x}^{2}}{{a}^{2} {x}^{2}}$

${f}_{3} = {x}^{3} - \frac{a {x}^{3} + {a}^{4}}{x + a} = \frac{{x}^{4} + a {x}^{3} - a {x}^{3} - {a}^{4}}{x + a} = \frac{{x}^{4} - {a}^{4}}{x + a}$

we have

$\frac{{f}_{1} {f}_{2}}{{f}_{3}} = \left(\frac{{x}^{2} + {a}^{2}}{x + a}\right) \left(\frac{{a}^{2} - {x}^{2}}{{a}^{2} {x}^{2}}\right) \left(\frac{x + a}{{x}^{4} - {a}^{4}}\right) = - \frac{1}{{a}^{2} {x}^{2}}$