Solve 1/31^3-4/31^2+8

$$\frac{ 1 }{ 3 } { 1 }^{ 3 } - \frac{ 4 }{ 3 } { 1 }^{ 2 } +8$$

Evaluate

$$7$$

Solution Steps

Calculate 1 to the power of 3 and get 1.

$$\frac{1}{3}\times 1-\frac{4}{3}\times 1^{2}+8$$

Multiply \frac{1}{3} and 1 to get \frac{1}{3}.

$$\frac{1}{3}-\frac{4}{3}\times 1^{2}+8$$

Calculate 1 to the power of 2 and get 1.

$$\frac{1}{3}-\frac{4}{3}\times 1+8$$

Multiply \frac{4}{3} and 1 to get \frac{4}{3}.

$$\frac{1}{3}-\frac{4}{3}+8$$

Since \frac{1}{3} and \frac{4}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-4}{3}+8$$

Subtract 4 from 1 to get -3.

$$\frac{-3}{3}+8$$

Divide -3 by 3 to get -1.

$$-1+8$$

Add -1 and 8 to get 7.

$$7$$

Factor

$$7$$