Solve [(6^{2}-5^2):11+11^0]^3-7^0

$$[ ( 6 ^ { 2 } - 5 ^ { 2 } ) : 11 + 11 ^ { 0 } ] ^ { 3 } - 7 ^ { 0 }$$

Evaluate

$$7$$

Solution Steps

Calculate 6 to the power of 2 and get 36.

$$\left(\frac{36-5^{2}}{11}+11^{0}\right)^{3}-7^{0}$$

Calculate 5 to the power of 2 and get 25.

$$\left(\frac{36-25}{11}+11^{0}\right)^{3}-7^{0}$$

Subtract 25 from 36 to get 11.

$$\left(\frac{11}{11}+11^{0}\right)^{3}-7^{0}$$

Divide 11 by 11 to get 1.

$$\left(1+11^{0}\right)^{3}-7^{0}$$

Calculate 11 to the power of 0 and get 1.

$$\left(1+1\right)^{3}-7^{0}$$

Add 1 and 1 to get 2.

$$2^{3}-7^{0}$$

Calculate 2 to the power of 3 and get 8.

$$8-7^{0}$$

Calculate 7 to the power of 0 and get 1.

$$8-1$$

Subtract 1 from 8 to get 7.

$$7$$

Factor

$$7$$