Solve [3^{6}:3^{6}+(2^{2})^2]^3:(2^2*5-3)^2+2*3-20

$$[ 3 ^ { 6 } : 3 ^ { 6 } + ( 2 ^ { 2 } ) ^ { 2 } ] ^ { 3 } : ( 2 ^ { 2 } \times 5 - 3 ) ^ { 2 } + 2 \times 3 - 20$$

Evaluate

$$3$$

Solution Steps

Divide 3^{6} by 3^{6} to get 1.

$$\frac{\left(1+\left(2^{2}\right)^{2}\right)^{3}}{\left(2^{2}\times 5-3\right)^{2}}+2\times 3-20$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{\left(1+2^{4}\right)^{3}}{\left(2^{2}\times 5-3\right)^{2}}+2\times 3-20$$

Calculate 2 to the power of 4 and get 16.

$$\frac{\left(1+16\right)^{3}}{\left(2^{2}\times 5-3\right)^{2}}+2\times 3-20$$

Add 1 and 16 to get 17.

$$\frac{17^{3}}{\left(2^{2}\times 5-3\right)^{2}}+2\times 3-20$$

Calculate 17 to the power of 3 and get 4913.

$$\frac{4913}{\left(2^{2}\times 5-3\right)^{2}}+2\times 3-20$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4913}{\left(4\times 5-3\right)^{2}}+2\times 3-20$$

Multiply 4 and 5 to get 20.

$$\frac{4913}{\left(20-3\right)^{2}}+2\times 3-20$$

Subtract 3 from 20 to get 17.

$$\frac{4913}{17^{2}}+2\times 3-20$$

Calculate 17 to the power of 2 and get 289.

$$\frac{4913}{289}+2\times 3-20$$

Divide 4913 by 289 to get 17.

$$17+2\times 3-20$$

Multiply 2 and 3 to get 6.

$$17+6-20$$

Add 17 and 6 to get 23.

$$23-20$$

Subtract 20 from 23 to get 3.

$$3$$

Factor

$$3$$