Solve sqrt[1]{2^{3}*3-6*(7*3-2*3^{2})+3^2*2^3-3-11^3:11^2}

$$\sqrt[ 1 ] { 2 ^ { 3 } \times 3 - 6 \times ( 7 \times 3 - 2 \times 3 ^ { 2 } ) + 3 ^ { 2 } \times 2 ^ { 3 } - 3 - 11 ^ { 3 } : 11 ^ { 2 } }$$

Evaluate

$$64$$

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 2 from 3 to get 1.

$$\sqrt[1]{2^{3}\times 3-6\left(7\times 3-2\times 3^{2}\right)+3^{2}\times 2^{3}-3-11^{1}}$$

Calculate 2 to the power of 3 and get 8.

$$\sqrt[1]{8\times 3-6\left(7\times 3-2\times 3^{2}\right)+3^{2}\times 2^{3}-3-11^{1}}$$

Multiply 8 and 3 to get 24.

$$\sqrt[1]{24-6\left(7\times 3-2\times 3^{2}\right)+3^{2}\times 2^{3}-3-11^{1}}$$

Multiply 7 and 3 to get 21.

$$\sqrt[1]{24-6\left(21-2\times 3^{2}\right)+3^{2}\times 2^{3}-3-11^{1}}$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt[1]{24-6\left(21-2\times 9\right)+3^{2}\times 2^{3}-3-11^{1}}$$

Multiply 2 and 9 to get 18.

$$\sqrt[1]{24-6\left(21-18\right)+3^{2}\times 2^{3}-3-11^{1}}$$

Subtract 18 from 21 to get 3.

$$\sqrt[1]{24-6\times 3+3^{2}\times 2^{3}-3-11^{1}}$$

Multiply 6 and 3 to get 18.

$$\sqrt[1]{24-18+3^{2}\times 2^{3}-3-11^{1}}$$

Subtract 18 from 24 to get 6.

$$\sqrt[1]{6+3^{2}\times 2^{3}-3-11^{1}}$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt[1]{6+9\times 2^{3}-3-11^{1}}$$

Calculate 2 to the power of 3 and get 8.

$$\sqrt[1]{6+9\times 8-3-11^{1}}$$

Multiply 9 and 8 to get 72.

$$\sqrt[1]{6+72-3-11^{1}}$$

Add 6 and 72 to get 78.

$$\sqrt[1]{78-3-11^{1}}$$

Subtract 3 from 78 to get 75.

$$\sqrt[1]{75-11^{1}}$$

Calculate 11 to the power of 1 and get 11.

$$\sqrt[1]{75-11}$$

Subtract 11 from 75 to get 64.

$$\sqrt[1]{64}$$

Calculate \sqrt[1]{64} and get 64.

$$64$$

Factor

$$2^{6}$$