Solve sqrt{9}-(3/2)^2]^2*sqrt{(sqrt{625}-12:sqrt[3]{216/125}})^2-3^2*sqrt[3]{729}

$$\sqrt { 9 } - ( \frac { 3 } { 2 } ) ^ { 2 } ] ^ { 2 } \cdot \sqrt { ( \sqrt { 625 } - 12 : \sqrt[ 3 ] { \frac { 216 } { 125 } } } ) ^ { 2 } - 3 ^ { 2 } \cdot \sqrt[ 3 ] { 729 }$$

Evaluate

$$\frac{3}{4}=0.75$$

Solution Steps

Calculate the square root of 9 and get 3.

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

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$3-\frac{9}{4}$$

Convert 3 to fraction \frac{12}{4}.

$$\frac{12}{4}-\frac{9}{4}$$

Since \frac{12}{4} and \frac{9}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{12-9}{4}$$

Subtract 9 from 12 to get 3.

$$\frac{3}{4}$$

Factor

$$\frac{3}{2 ^ {2}} = 0.75$$