Solve (5^{4}:5^{3}-14:7)^4:(3^2*2-3*5)^2-(2^2*3-2*5)

$$( 5 ^ { 4 } : 5 ^ { 3 } - 14 : 7 ) ^ { 4 } : ( 3 ^ { 2 } \times 2 - 3 \times 5 ) ^ { 2 } - ( 2 ^ { 2 } \times 3 - 2 \times 5 )$$

Evaluate

$$7$$

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 3 from 4 to get 1.

$$\frac{\left(5^{1}-\frac{14}{7}\right)^{4}}{\left(3^{2}\times 2-3\times 5\right)^{2}}-\left(2^{2}\times 3-2\times 5\right)$$

Calculate 5 to the power of 1 and get 5.

$$\frac{\left(5-\frac{14}{7}\right)^{4}}{\left(3^{2}\times 2-3\times 5\right)^{2}}-\left(2^{2}\times 3-2\times 5\right)$$

Divide 14 by 7 to get 2.

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

Subtract 2 from 5 to get 3.

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

Calculate 3 to the power of 4 and get 81.

$$\frac{81}{\left(3^{2}\times 2-3\times 5\right)^{2}}-\left(2^{2}\times 3-2\times 5\right)$$

Calculate 3 to the power of 2 and get 9.

$$\frac{81}{\left(9\times 2-3\times 5\right)^{2}}-\left(2^{2}\times 3-2\times 5\right)$$

Multiply 9 and 2 to get 18.

$$\frac{81}{\left(18-3\times 5\right)^{2}}-\left(2^{2}\times 3-2\times 5\right)$$

Multiply 3 and 5 to get 15.

$$\frac{81}{\left(18-15\right)^{2}}-\left(2^{2}\times 3-2\times 5\right)$$

Subtract 15 from 18 to get 3.

$$\frac{81}{3^{2}}-\left(2^{2}\times 3-2\times 5\right)$$

Calculate 3 to the power of 2 and get 9.

$$\frac{81}{9}-\left(2^{2}\times 3-2\times 5\right)$$

Divide 81 by 9 to get 9.

$$9-\left(2^{2}\times 3-2\times 5\right)$$

Calculate 2 to the power of 2 and get 4.

$$9-\left(4\times 3-2\times 5\right)$$

Multiply 4 and 3 to get 12.

$$9-\left(12-2\times 5\right)$$

Multiply 2 and 5 to get 10.

$$9-\left(12-10\right)$$

Subtract 10 from 12 to get 2.

$$9-2$$

Subtract 2 from 9 to get 7.

$$7$$

Factor

$$7$$