Solve sqrt{2^3*3-6*(7*3-2*3^2)+3^2*2^3-{11}^3div{11}^2}

$$\sqrt{ { 2 }^{ 3 } \times 3-6 \times (7 \times 3-2 \times { 3 }^{ 2 } )+ { 3 }^{ 2 } \times { 2 }^{ 3 } - { 11 }^{ 3 } \div { 11 }^{ 2 } }$$

Evaluate

$$\sqrt{67}\approx 8.185352772$$

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.

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

Calculate 2 to the power of 3 and get 8.

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

Multiply 8 and 3 to get 24.

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

Multiply 7 and 3 to get 21.

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

Calculate 3 to the power of 2 and get 9.

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

Multiply 2 and 9 to get 18.

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

Subtract 18 from 21 to get 3.

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

Multiply 6 and 3 to get 18.

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

Subtract 18 from 24 to get 6.

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

Calculate 3 to the power of 2 and get 9.

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

Calculate 2 to the power of 3 and get 8.

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

Multiply 9 and 8 to get 72.

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

Add 6 and 72 to get 78.

$$\sqrt{78-11^{1}}$$

Calculate 11 to the power of 1 and get 11.

$$\sqrt{78-11}$$

Subtract 11 from 78 to get 67.

$$\sqrt{67}$$