Solve frac{a^{3}+b^{3}}{2a^{2}-ab-3b^{2}}diva^2-ab+b^2/4a^3-9ab^2

$$\frac { a ^ { 3 } + b ^ { 3 } } { 2 a ^ { 2 } - a b - 3 b ^ { 2 } } \div \frac { a ^ { 2 } - a b + b ^ { 2 } } { 4 a ^ { 3 } - 9 a b ^ { 2 } }$$

Evaluate

$$a\left(2a+3b\right)$$

Short Solution Steps

Divide \frac{a^{3}+b^{3}}{2a^{2}-ab-3b^{2}} by \frac{a^{2}-ab+b^{2}}{4a^{3}-9ab^{2}} by multiplying \frac{a^{3}+b^{3}}{2a^{2}-ab-3b^{2}} by the reciprocal of \frac{a^{2}-ab+b^{2}}{4a^{3}-9ab^{2}}.

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

Factor the expressions that are not already factored.

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

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

$$a\left(2a+3b\right)$$

Expand the expression.

$$2a^{2}+3ab$$

Expand

$$2a^{2}+3ab$$

Short Solution Steps

Divide \frac{a^{3}+b^{3}}{2a^{2}-ab-3b^{2}} by \frac{a^{2}-ab+b^{2}}{4a^{3}-9ab^{2}} by multiplying \frac{a^{3}+b^{3}}{2a^{2}-ab-3b^{2}} by the reciprocal of \frac{a^{2}-ab+b^{2}}{4a^{3}-9ab^{2}}.

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

Factor the expressions that are not already factored.

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

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

$$a\left(2a+3b\right)$$

Expand the expression.

$$2a^{2}+3ab$$