Solve 2^{3}*(3^{2}-3^1)-4*[(2^3+3^2)*2-3^3]

$$2 ^ { 3 } \times ( 3 ^ { 2 } - 3 ^ { 1 } ) - 4 \times [ ( 2 ^ { 3 } + 3 ^ { 2 } ) \times 2 - 3 ^ { 3 } ]$$

Evaluate

$$20$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$8\left(3^{2}-3^{1}\right)-4\left(\left(2^{3}+3^{2}\right)\times 2-3^{3}\right)$$

Calculate 3 to the power of 2 and get 9.

$$8\left(9-3^{1}\right)-4\left(\left(2^{3}+3^{2}\right)\times 2-3^{3}\right)$$

Calculate 3 to the power of 1 and get 3.

$$8\left(9-3\right)-4\left(\left(2^{3}+3^{2}\right)\times 2-3^{3}\right)$$

Subtract 3 from 9 to get 6.

$$8\times 6-4\left(\left(2^{3}+3^{2}\right)\times 2-3^{3}\right)$$

Multiply 8 and 6 to get 48.

$$48-4\left(\left(2^{3}+3^{2}\right)\times 2-3^{3}\right)$$

Calculate 2 to the power of 3 and get 8.

$$48-4\left(\left(8+3^{2}\right)\times 2-3^{3}\right)$$

Calculate 3 to the power of 2 and get 9.

$$48-4\left(\left(8+9\right)\times 2-3^{3}\right)$$

Add 8 and 9 to get 17.

$$48-4\left(17\times 2-3^{3}\right)$$

Multiply 17 and 2 to get 34.

$$48-4\left(34-3^{3}\right)$$

Calculate 3 to the power of 3 and get 27.

$$48-4\left(34-27\right)$$

Subtract 27 from 34 to get 7.

$$48-4\times 7$$

Multiply 4 and 7 to get 28.

$$48-28$$

Subtract 28 from 48 to get 20.

$$20$$

Factor

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