Solve 3(20+4)/2(9)-4^2*sqrt[3]{5(8)+41/12-3(16-13)}=

$$\frac { 3 ( 20 + 4 ) } { 2 ( 9 ) - 4 ^ { 2 } } \cdot \sqrt[ 3 ] { \frac { 5 ( 8 ) + 41 } { 12 - 3 ( 16 - 13 ) } } =$$

Evaluate

$$108$$

Solution Steps

Add 20 and 4 to get 24.

$$\frac{3\times 24}{2\times 9-4^{2}}\sqrt[3]{\frac{5\times 8+41}{12-3\left(16-13\right)}}$$

Multiply 3 and 24 to get 72.

$$\frac{72}{2\times 9-4^{2}}\sqrt[3]{\frac{5\times 8+41}{12-3\left(16-13\right)}}$$

Multiply 2 and 9 to get 18.

$$\frac{72}{18-4^{2}}\sqrt[3]{\frac{5\times 8+41}{12-3\left(16-13\right)}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{72}{18-16}\sqrt[3]{\frac{5\times 8+41}{12-3\left(16-13\right)}}$$

Subtract 16 from 18 to get 2.

$$\frac{72}{2}\sqrt[3]{\frac{5\times 8+41}{12-3\left(16-13\right)}}$$

Divide 72 by 2 to get 36.

$$36\sqrt[3]{\frac{5\times 8+41}{12-3\left(16-13\right)}}$$

Multiply 5 and 8 to get 40.

$$36\sqrt[3]{\frac{40+41}{12-3\left(16-13\right)}}$$

Add 40 and 41 to get 81.

$$36\sqrt[3]{\frac{81}{12-3\left(16-13\right)}}$$

Subtract 13 from 16 to get 3.

$$36\sqrt[3]{\frac{81}{12-3\times 3}}$$

Multiply 3 and 3 to get 9.

$$36\sqrt[3]{\frac{81}{12-9}}$$

Subtract 9 from 12 to get 3.

$$36\sqrt[3]{\frac{81}{3}}$$

Divide 81 by 3 to get 27.

$$36\sqrt[3]{27}$$

Calculate \sqrt[3]{27} and get 3.

$$36\times 3$$

Multiply 36 and 3 to get 108.

$$108$$

Factor

$$2^{2}\times 3^{3}$$