Solve (2^{2}+5^{3}*5^{4}:5^{6})^{2}:(4^3:4^2-4^0)^4

$$( 2 ^ { 2 } + 5 ^ { 3 } \cdot 5 ^ { 4 } : 5 ^ { 6 } ) ^ { 2 } : ( 4 ^ { 3 } : 4 ^ { 2 } - 4 ^ { 0 } ) ^ { 4 }$$

Evaluate

$$1$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 3 and 4 to get 7.

$$\frac{\left(2^{2}+\frac{5^{7}}{5^{6}}\right)^{2}}{\left(\frac{4^{3}}{4^{2}}-4^{0}\right)^{4}}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 6 from 7 to get 1.

$$\frac{\left(2^{2}+5^{1}\right)^{2}}{\left(\frac{4^{3}}{4^{2}}-4^{0}\right)^{4}}$$

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

$$\frac{\left(2^{2}+5^{1}\right)^{2}}{\left(4^{1}-4^{0}\right)^{4}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(4+5^{1}\right)^{2}}{\left(4^{1}-4^{0}\right)^{4}}$$

Calculate 5 to the power of 1 and get 5.

$$\frac{\left(4+5\right)^{2}}{\left(4^{1}-4^{0}\right)^{4}}$$

Add 4 and 5 to get 9.

$$\frac{9^{2}}{\left(4^{1}-4^{0}\right)^{4}}$$

Calculate 9 to the power of 2 and get 81.

$$\frac{81}{\left(4^{1}-4^{0}\right)^{4}}$$

Calculate 4 to the power of 1 and get 4.

$$\frac{81}{\left(4-4^{0}\right)^{4}}$$

Calculate 4 to the power of 0 and get 1.

$$\frac{81}{\left(4-1\right)^{4}}$$

Subtract 1 from 4 to get 3.

$$\frac{81}{3^{4}}$$

Calculate 3 to the power of 4 and get 81.

$$\frac{81}{81}$$

Divide 81 by 81 to get 1.

$$1$$

Factor

$$1$$