Solve ((a)^(2)+ab-2(b)^(2))/((a)^(2)-2ab-3(b)^(2))\times((a)^(2)-(b)^(2))/((a)^(2)+2(b)^(2))\div((a)^(2)-2ab+(b)^(2))/((a)^(2)-3ab)

Question

$$\frac{ { a }^{ 2 } +ab-2 { b }^{ 2 } }{ { a }^{ 2 } -2ab-3 { b }^{ 2 } } \times \frac{ { a }^{ 2 } - { b }^{ 2 } }{ { a }^{ 2 } +2 { b }^{ 2 } } \div \frac{ { a }^{ 2 } -2ab+ { b }^{ 2 } }{ { a }^{ 2 } -3ab }$$

Evaluate

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

Short Solution Steps

Multiply $\frac{a^{2}+ab-2b^{2}}{a^{2}-2ab-3b^{2}}$ times $\frac{a^{2}-b^{2}}{a^{2}+2b^{2}}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{\left(a^{2}+ab-2b^{2}\right)\left(a^{2}-b^{2}\right)}{\left(a^{2}-2ab-3b^{2}\right)\left(a^{2}+2b^{2}\right)}}{\frac{a^{2}-2ab+b^{2}}{a^{2}-3ab}}$$

Divide $\frac{\left(a^{2}+ab-2b^{2}\right)\left(a^{2}-b^{2}\right)}{\left(a^{2}-2ab-3b^{2}\right)\left(a^{2}+2b^{2}\right)}$ by $\frac{a^{2}-2ab+b^{2}}{a^{2}-3ab}$ by multiplying $\frac{\left(a^{2}+ab-2b^{2}\right)\left(a^{2}-b^{2}\right)}{\left(a^{2}-2ab-3b^{2}\right)\left(a^{2}+2b^{2}\right)}$ by the reciprocal of $\frac{a^{2}-2ab+b^{2}}{a^{2}-3ab}$.

$$\frac{\left(a^{2}+ab-2b^{2}\right)\left(a^{2}-b^{2}\right)\left(a^{2}-3ab\right)}{\left(a^{2}-2ab-3b^{2}\right)\left(a^{2}+2b^{2}\right)\left(a^{2}-2ab+b^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{a\left(a+b\right)\left(a-3b\right)\left(a+2b\right)\left(a-b\right)^{2}}{\left(a+b\right)\left(a-3b\right)\left(a-b\right)^{2}\left(a^{2}+2b^{2}\right)}$$

Cancel out $\left(a+b\right)\left(a-3b\right)\left(a-b\right)^{2}$ in both numerator and denominator.

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

Expand the expression.

$$\frac{a^{2}+2ab}{a^{2}+2b^{2}}$$

Show Solution

Expand

$\frac{a^{2}+2ab}{a^{2}+2b^{2}}$

Short Solution Steps

Multiply $\frac{a^{2}+ab-2b^{2}}{a^{2}-2ab-3b^{2}}$ times $\frac{a^{2}-b^{2}}{a^{2}+2b^{2}}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{\left(a^{2}+ab-2b^{2}\right)\left(a^{2}-b^{2}\right)}{\left(a^{2}-2ab-3b^{2}\right)\left(a^{2}+2b^{2}\right)}}{\frac{a^{2}-2ab+b^{2}}{a^{2}-3ab}}$$

$$\frac{\left(a^{2}+ab-2b^{2}\right)\left(a^{2}-b^{2}\right)\left(a^{2}-3ab\right)}{\left(a^{2}-2ab-3b^{2}\right)\left(a^{2}+2b^{2}\right)\left(a^{2}-2ab+b^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{a\left(a+b\right)\left(a-3b\right)\left(a+2b\right)\left(a-b\right)^{2}}{\left(a+b\right)\left(a-3b\right)\left(a-b\right)^{2}\left(a^{2}+2b^{2}\right)}$$

Cancel out $\left(a+b\right)\left(a-3b\right)\left(a-b\right)^{2}$ in both numerator and denominator.

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

Expand the expression.

$$\frac{a^{2}+2ab}{a^{2}+2b^{2}}$$

Show Solution

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