Solve sqrt{{(1/6)^{2}+[(1/3)^{2}-(1/3-1/6)^{2}]}*{[frac{1}{2}+(10^circ-frac{1}{3}-frac{4}{6})^4]-frac{2}{3^2}-(frac{1}{2}-frac{1}{3})^2}}

$$\sqrt { \{ ( \frac { 1 } { 6 } ) ^ { 2 } + [ ( \frac { 1 } { 3 } ) ^ { 2 } - ( \frac { 1 } { 3 } - \frac { 1 } { 6 } ) ^ { 2 } ] \} \times \{ [ \frac { 1 } { 2 } + ( 10 ^ { \circ } - \frac { 1 } { 3 } - \frac { 4 } { 6 } ) ^ { 4 } ] - \frac { 2 } { 3 ^ { 2 } } - ( \frac { 1 } { 2 } - \frac { 1 } { 3 } ) ^ { 2 } \} }$$

Evaluate

$$\frac{\sqrt{26245}}{6}\approx 27.000514398$$

Solution Steps

Calculate \frac{1}{6} to the power of 2 and get \frac{1}{36}.

$$\sqrt{\left(\frac{1}{36}+\left(\frac{1}{3}\right)^{2}-\left(\frac{1}{3}-\frac{1}{6}\right)^{2}\right)\left(\frac{1}{2}+\left(10-\frac{1}{3}-\frac{4}{6}\right)^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$\sqrt{\left(\frac{1}{36}+\frac{1}{9}-\left(\frac{1}{3}-\frac{1}{6}\right)^{2}\right)\left(\frac{1}{2}+\left(10-\frac{1}{3}-\frac{4}{6}\right)^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Subtract \frac{1}{6} from \frac{1}{3} to get \frac{1}{6}.

$$\sqrt{\left(\frac{1}{36}+\frac{1}{9}-\left(\frac{1}{6}\right)^{2}\right)\left(\frac{1}{2}+\left(10-\frac{1}{3}-\frac{4}{6}\right)^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Calculate \frac{1}{6} to the power of 2 and get \frac{1}{36}.

$$\sqrt{\left(\frac{1}{36}+\frac{1}{9}-\frac{1}{36}\right)\left(\frac{1}{2}+\left(10-\frac{1}{3}-\frac{4}{6}\right)^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Subtract \frac{1}{36} from \frac{1}{9} to get \frac{1}{12}.

$$\sqrt{\left(\frac{1}{36}+\frac{1}{12}\right)\left(\frac{1}{2}+\left(10-\frac{1}{3}-\frac{4}{6}\right)^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Add \frac{1}{36} and \frac{1}{12} to get \frac{1}{9}.

$$\sqrt{\frac{1}{9}\left(\frac{1}{2}+\left(10-\frac{1}{3}-\frac{4}{6}\right)^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Subtract \frac{1}{3} from 10 to get \frac{29}{3}.

$$\sqrt{\frac{1}{9}\left(\frac{1}{2}+\left(\frac{29}{3}-\frac{4}{6}\right)^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$\sqrt{\frac{1}{9}\left(\frac{1}{2}+\left(\frac{29}{3}-\frac{2}{3}\right)^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Subtract \frac{2}{3} from \frac{29}{3} to get 9.

$$\sqrt{\frac{1}{9}\left(\frac{1}{2}+9^{4}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Calculate 9 to the power of 4 and get 6561.

$$\sqrt{\frac{1}{9}\left(\frac{1}{2}+6561-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Add \frac{1}{2} and 6561 to get \frac{13123}{2}.

$$\sqrt{\frac{1}{9}\left(\frac{13123}{2}-\frac{2}{3^{2}}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt{\frac{1}{9}\left(\frac{13123}{2}-\frac{2}{9}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Subtract \frac{2}{9} from \frac{13123}{2} to get \frac{118103}{18}.

$$\sqrt{\frac{1}{9}\left(\frac{118103}{18}-\left(\frac{1}{2}-\frac{1}{3}\right)^{2}\right)}$$

Subtract \frac{1}{3} from \frac{1}{2} to get \frac{1}{6}.

$$\sqrt{\frac{1}{9}\left(\frac{118103}{18}-\left(\frac{1}{6}\right)^{2}\right)}$$

Calculate \frac{1}{6} to the power of 2 and get \frac{1}{36}.

$$\sqrt{\frac{1}{9}\left(\frac{118103}{18}-\frac{1}{36}\right)}$$

Subtract \frac{1}{36} from \frac{118103}{18} to get \frac{26245}{4}.

$$\sqrt{\frac{1}{9}\times \frac{26245}{4}}$$

Multiply \frac{1}{9} and \frac{26245}{4} to get \frac{26245}{36}.

$$\sqrt{\frac{26245}{36}}$$

Rewrite the square root of the division \sqrt{\frac{26245}{36}} as the division of square roots \frac{\sqrt{26245}}{\sqrt{36}}.

$$\frac{\sqrt{26245}}{\sqrt{36}}$$

Calculate the square root of 36 and get 6.

$$\frac{\sqrt{26245}}{6}$$