Solve {[(a+b)^3-(a-b)^3]:(6b)}:(-3/2)

$$\{ [ ( a + b ) ^ { 3 } - ( a - b ) ^ { 3 } ] : ( 6 b ) \} : ( - \frac { 3 } { 2 } )$$

Evaluate

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

Short Solution Steps

Divide \frac{\left(a+b\right)^{3}-\left(a-b\right)^{3}}{6b} by -\frac{3}{2} by multiplying \frac{\left(a+b\right)^{3}-\left(a-b\right)^{3}}{6b} by the reciprocal of -\frac{3}{2}.

$$\frac{\left(\left(a+b\right)^{3}-\left(a-b\right)^{3}\right)\times 2}{6b\left(-3\right)}$$

Cancel out 2 in both numerator and denominator.

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

Multiply -3 and 3 to get -9.

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

Factor the expressions that are not already factored.

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

Cancel out b in both numerator and denominator.

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

Expand the expression.

$$\frac{6a^{2}+2b^{2}}{-9}$$

Expand

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

Short Solution Steps

Divide \frac{\left(a+b\right)^{3}-\left(a-b\right)^{3}}{6b} by -\frac{3}{2} by multiplying \frac{\left(a+b\right)^{3}-\left(a-b\right)^{3}}{6b} by the reciprocal of -\frac{3}{2}.

$$\frac{\left(\left(a+b\right)^{3}-\left(a-b\right)^{3}\right)\times 2}{6b\left(-3\right)}$$

Cancel out 2 in both numerator and denominator.

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

Multiply -3 and 3 to get -9.

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

Factor the expressions that are not already factored.

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

Cancel out b in both numerator and denominator.

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

Expand the expression.

$$\frac{6a^{2}+2b^{2}}{-9}$$