Solve h(u)=(9+1/u)(u^3-1/8)

$$h ( u ) = ( 9 + \frac { 1 } { u } ) ( u ^ { 3 } - \frac { 1 } { 8 } )$$

Evaluate

$$9u^{3}+u^{2}-\frac{9}{8}-\frac{1}{8u}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{u}{u}.

$$\left(\frac{9u}{u}+\frac{1}{u}\right)\left(u^{3}-\frac{1}{8}\right)$$

Since \frac{9u}{u} and \frac{1}{u} have the same denominator, add them by adding their numerators.

$$\frac{9u+1}{u}\left(u^{3}-\frac{1}{8}\right)$$

Use the distributive property to multiply \frac{9u+1}{u} by u^{3}-\frac{1}{8}.

$$\frac{9u+1}{u}u^{3}-\frac{1}{8}\times \frac{9u+1}{u}$$

Express \frac{9u+1}{u}u^{3} as a single fraction.

$$\frac{\left(9u+1\right)u^{3}}{u}-\frac{1}{8}\times \frac{9u+1}{u}$$

Cancel out u in both numerator and denominator.

$$\left(9u+1\right)u^{2}-\frac{1}{8}\times \frac{9u+1}{u}$$

Use the distributive property to multiply 9u+1 by u^{2}.

$$9u^{3}+u^{2}-\frac{1}{8}\times \frac{9u+1}{u}$$

Multiply -\frac{1}{8} times \frac{9u+1}{u} by multiplying numerator times numerator and denominator times denominator.

$$9u^{3}+u^{2}+\frac{-\left(9u+1\right)}{8u}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9u^{3}+u^{2} times \frac{8u}{8u}.

$$\frac{\left(9u^{3}+u^{2}\right)\times 8u}{8u}+\frac{-\left(9u+1\right)}{8u}$$

Since \frac{\left(9u^{3}+u^{2}\right)\times 8u}{8u} and \frac{-\left(9u+1\right)}{8u} have the same denominator, add them by adding their numerators.

$$\frac{\left(9u^{3}+u^{2}\right)\times 8u-\left(9u+1\right)}{8u}$$

Do the multiplications in \left(9u^{3}+u^{2}\right)\times 8u-\left(9u+1\right).

$$\frac{72u^{4}+8u^{3}-9u-1}{8u}$$

Expand

$$9u^{3}+u^{2}-\frac{9}{8}-\frac{1}{8u}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{u}{u}.

$$\left(\frac{9u}{u}+\frac{1}{u}\right)\left(u^{3}-\frac{1}{8}\right)$$

Since \frac{9u}{u} and \frac{1}{u} have the same denominator, add them by adding their numerators.

$$\frac{9u+1}{u}\left(u^{3}-\frac{1}{8}\right)$$

Use the distributive property to multiply \frac{9u+1}{u} by u^{3}-\frac{1}{8}.

$$\frac{9u+1}{u}u^{3}-\frac{1}{8}\times \frac{9u+1}{u}$$

Express \frac{9u+1}{u}u^{3} as a single fraction.

$$\frac{\left(9u+1\right)u^{3}}{u}-\frac{1}{8}\times \frac{9u+1}{u}$$

Cancel out u in both numerator and denominator.

$$\left(9u+1\right)u^{2}-\frac{1}{8}\times \frac{9u+1}{u}$$

Use the distributive property to multiply 9u+1 by u^{2}.

$$9u^{3}+u^{2}-\frac{1}{8}\times \frac{9u+1}{u}$$

Multiply -\frac{1}{8} times \frac{9u+1}{u} by multiplying numerator times numerator and denominator times denominator.

$$9u^{3}+u^{2}+\frac{-\left(9u+1\right)}{8u}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9u^{3}+u^{2} times \frac{8u}{8u}.

$$\frac{\left(9u^{3}+u^{2}\right)\times 8u}{8u}+\frac{-\left(9u+1\right)}{8u}$$

Since \frac{\left(9u^{3}+u^{2}\right)\times 8u}{8u} and \frac{-\left(9u+1\right)}{8u} have the same denominator, add them by adding their numerators.

$$\frac{\left(9u^{3}+u^{2}\right)\times 8u-\left(9u+1\right)}{8u}$$

Do the multiplications in \left(9u^{3}+u^{2}\right)\times 8u-\left(9u+1\right).

$$\frac{72u^{4}+8u^{3}-9u-1}{8u}$$