Solve (4^2-2^3)div(31/8+7/8)^2

$$( 4 ^ { 2 } - 2 ^ { 3 } ) \div ( 3 \frac { 1 } { 8 } + \frac { 7 } { 8 } ) ^ { 2 }$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$\frac{16-2^{3}}{\left(\frac{3\times 8+1}{8}+\frac{7}{8}\right)^{2}}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{16-8}{\left(\frac{3\times 8+1}{8}+\frac{7}{8}\right)^{2}}$$

Subtract 8 from 16 to get 8.

$$\frac{8}{\left(\frac{3\times 8+1}{8}+\frac{7}{8}\right)^{2}}$$

Multiply 3 and 8 to get 24.

$$\frac{8}{\left(\frac{24+1}{8}+\frac{7}{8}\right)^{2}}$$

Add 24 and 1 to get 25.

$$\frac{8}{\left(\frac{25}{8}+\frac{7}{8}\right)^{2}}$$

Add \frac{25}{8} and \frac{7}{8} to get 4.

$$\frac{8}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$