Solve {3^2*2^3-(63:9)+[sqrt{121}*sqrt[3]{27}]-2}=

$$\{ 3 ^ { 2 } \cdot 2 ^ { 3 } - ( 63 : 9 ) + [ \sqrt { 121 } \cdot \sqrt[ 3 ] { 27 } ] - 2 \} =$$

Evaluate

$$96$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$9\times 2^{3}-\frac{63}{9}+\sqrt{121}\sqrt[3]{27}-2$$

Calculate 2 to the power of 3 and get 8.

$$9\times 8-\frac{63}{9}+\sqrt{121}\sqrt[3]{27}-2$$

Multiply 9 and 8 to get 72.

$$72-\frac{63}{9}+\sqrt{121}\sqrt[3]{27}-2$$

Divide 63 by 9 to get 7.

$$72-7+\sqrt{121}\sqrt[3]{27}-2$$

Subtract 7 from 72 to get 65.

$$65+\sqrt{121}\sqrt[3]{27}-2$$

Calculate the square root of 121 and get 11.

$$65+11\sqrt[3]{27}-2$$

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

$$65+11\times 3-2$$

Multiply 11 and 3 to get 33.

$$65+33-2$$

Add 65 and 33 to get 98.

$$98-2$$

Subtract 2 from 98 to get 96.

$$96$$

Factor

$$2^{5}\times 3$$