Solve (3*2^3)^2-23^2+60:3+3^3*2=

$$( 3 \cdot 2 ^ { 3 } ) ^ { 2 } - 23 ^ { 2 } + 60 : 3 + 3 ^ { 3 } \cdot 2 =$$

Evaluate

$$121$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$\left(3\times 8\right)^{2}-23^{2}+\frac{60}{3}+3^{3}\times 2$$

Multiply 3 and 8 to get 24.

$$24^{2}-23^{2}+\frac{60}{3}+3^{3}\times 2$$

Calculate 24 to the power of 2 and get 576.

$$576-23^{2}+\frac{60}{3}+3^{3}\times 2$$

Calculate 23 to the power of 2 and get 529.

$$576-529+\frac{60}{3}+3^{3}\times 2$$

Subtract 529 from 576 to get 47.

$$47+\frac{60}{3}+3^{3}\times 2$$

Divide 60 by 3 to get 20.

$$47+20+3^{3}\times 2$$

Add 47 and 20 to get 67.

$$67+3^{3}\times 2$$

Calculate 3 to the power of 3 and get 27.

$$67+27\times 2$$

Multiply 27 and 2 to get 54.

$$67+54$$

Add 67 and 54 to get 121.

$$121$$

Factor

$$11^{2}$$