Solve sqrt{25}-6*3div3^2+sqrt[5]{-32}

$$\sqrt { 25 } - 6 \times 3 \div 3 ^ { 2 } + \sqrt[ 5 ] { - 32 }$$

Evaluate

$$1$$

Solution Steps

Calculate the square root of 25 and get 5.

$$5-\frac{6\times 3}{3^{2}}+\sqrt[5]{-32}$$

Cancel out 3 in both numerator and denominator.

$$5-\frac{6}{3}+\sqrt[5]{-32}$$

Cancel out 3 in both numerator and denominator.

$$5-2+\sqrt[5]{-32}$$

Subtract 2 from 5 to get 3.

$$3+\sqrt[5]{-32}$$

Calculate \sqrt[5]{-32} and get -2.

$$3-2$$

Subtract 2 from 3 to get 1.

$$1$$

Factor

$$1$$