Solve (sqrt{1/9}-[(1/3)^2:(1/2)^2*1.5-(0.5+22/3)+0.5]^3+frac{2}{3})

$$( \sqrt { \frac { 1 } { 9 } } - [ ( \frac { 1 } { 3 } ) ^ { 2 } : ( \frac { 1 } { 2 } ) ^ { 2 } \times 1.5 - ( 0.5 + 2 \frac { 2 } { 3 } ) + 0.5 ] ^ { 3 } + \frac { 2 } { 3 } )$$

Evaluate

$$9$$

Solution Steps

Rewrite the square root of the division \frac{1}{9} as the division of square roots \frac{\sqrt{1}}{\sqrt{9}}. Take the square root of both numerator and denominator.

$$\frac{1}{3}-\left(\frac{\left(\frac{1}{3}\right)^{2}}{\left(\frac{1}{2}\right)^{2}}\times 1.5-\left(0.5+\frac{2\times 3+2}{3}\right)+0.5\right)^{3}+\frac{2}{3}$$

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

$$\frac{1}{3}-\left(\frac{\frac{1}{9}}{\left(\frac{1}{2}\right)^{2}}\times 1.5-\left(0.5+\frac{2\times 3+2}{3}\right)+0.5\right)^{3}+\frac{2}{3}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{3}-\left(\frac{\frac{1}{9}}{\frac{1}{4}}\times 1.5-\left(0.5+\frac{2\times 3+2}{3}\right)+0.5\right)^{3}+\frac{2}{3}$$

Divide \frac{1}{9} by \frac{1}{4} by multiplying \frac{1}{9} by the reciprocal of \frac{1}{4}.

$$\frac{1}{3}-\left(\frac{1}{9}\times 4\times 1.5-\left(0.5+\frac{2\times 3+2}{3}\right)+0.5\right)^{3}+\frac{2}{3}$$

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

$$\frac{1}{3}-\left(\frac{4}{9}\times 1.5-\left(0.5+\frac{2\times 3+2}{3}\right)+0.5\right)^{3}+\frac{2}{3}$$

Multiply \frac{4}{9} and 1.5 to get \frac{2}{3}.

$$\frac{1}{3}-\left(\frac{2}{3}-\left(0.5+\frac{2\times 3+2}{3}\right)+0.5\right)^{3}+\frac{2}{3}$$

Multiply 2 and 3 to get 6.

$$\frac{1}{3}-\left(\frac{2}{3}-\left(0.5+\frac{6+2}{3}\right)+0.5\right)^{3}+\frac{2}{3}$$

Add 6 and 2 to get 8.

$$\frac{1}{3}-\left(\frac{2}{3}-\left(0.5+\frac{8}{3}\right)+0.5\right)^{3}+\frac{2}{3}$$

Add 0.5 and \frac{8}{3} to get \frac{19}{6}.

$$\frac{1}{3}-\left(\frac{2}{3}-\frac{19}{6}+0.5\right)^{3}+\frac{2}{3}$$

Subtract \frac{19}{6} from \frac{2}{3} to get -\frac{5}{2}.

$$\frac{1}{3}-\left(-\frac{5}{2}+0.5\right)^{3}+\frac{2}{3}$$

Add -\frac{5}{2} and 0.5 to get -2.

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

Calculate -2 to the power of 3 and get -8.

$$\frac{1}{3}-\left(-8\right)+\frac{2}{3}$$

The opposite of -8 is 8.

$$\frac{1}{3}+8+\frac{2}{3}$$

Add \frac{1}{3} and 8 to get \frac{25}{3}.

$$\frac{25}{3}+\frac{2}{3}$$

Add \frac{25}{3} and \frac{2}{3} to get 9.

$$9$$

Factor

$$3^{2}$$