Solve h(u)=(9+1/u)(u^4-1/3)

$$h ( u ) = ( 9 + \frac { 1 } { u } ) ( u ^ { 4 } - \frac { 1 } { 3 } )$$

Evaluate

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

Short Solution Steps

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

$$\left(\frac{9u}{u}+\frac{1}{u}\right)\left(u^{4}-\frac{1}{3}\right)$$

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

$$\frac{9u+1}{u}\left(u^{4}-\frac{1}{3}\right)$$

Use the distributive property to multiply \frac{9u+1}{u} by u^{4}-\frac{1}{3}.

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

Express \frac{9u+1}{u}u^{4} as a single fraction.

$$\frac{\left(9u+1\right)u^{4}}{u}-\frac{1}{3}\times \frac{9u+1}{u}$$

Cancel out u in both numerator and denominator.

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

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

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

Multiply -\frac{1}{3} times \frac{9u+1}{u} by multiplying numerator times numerator and denominator times denominator.

$$9u^{4}+u^{3}+\frac{-\left(9u+1\right)}{3u}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9u^{4}+u^{3} times \frac{3u}{3u}.

$$\frac{\left(9u^{4}+u^{3}\right)\times 3u}{3u}+\frac{-\left(9u+1\right)}{3u}$$

Since \frac{\left(9u^{4}+u^{3}\right)\times 3u}{3u} and \frac{-\left(9u+1\right)}{3u} have the same denominator, add them by adding their numerators.

$$\frac{\left(9u^{4}+u^{3}\right)\times 3u-\left(9u+1\right)}{3u}$$

Do the multiplications in \left(9u^{4}+u^{3}\right)\times 3u-\left(9u+1\right).

$$\frac{27u^{5}+3u^{4}-9u-1}{3u}$$

Expand

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

Short Solution Steps

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

$$\left(\frac{9u}{u}+\frac{1}{u}\right)\left(u^{4}-\frac{1}{3}\right)$$

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

$$\frac{9u+1}{u}\left(u^{4}-\frac{1}{3}\right)$$

Use the distributive property to multiply \frac{9u+1}{u} by u^{4}-\frac{1}{3}.

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

Express \frac{9u+1}{u}u^{4} as a single fraction.

$$\frac{\left(9u+1\right)u^{4}}{u}-\frac{1}{3}\times \frac{9u+1}{u}$$

Cancel out u in both numerator and denominator.

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

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

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

Multiply -\frac{1}{3} times \frac{9u+1}{u} by multiplying numerator times numerator and denominator times denominator.

$$9u^{4}+u^{3}+\frac{-\left(9u+1\right)}{3u}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9u^{4}+u^{3} times \frac{3u}{3u}.

$$\frac{\left(9u^{4}+u^{3}\right)\times 3u}{3u}+\frac{-\left(9u+1\right)}{3u}$$

Since \frac{\left(9u^{4}+u^{3}\right)\times 3u}{3u} and \frac{-\left(9u+1\right)}{3u} have the same denominator, add them by adding their numerators.

$$\frac{\left(9u^{4}+u^{3}\right)\times 3u-\left(9u+1\right)}{3u}$$

Do the multiplications in \left(9u^{4}+u^{3}\right)\times 3u-\left(9u+1\right).

$$\frac{27u^{5}+3u^{4}-9u-1}{3u}$$