Solve [(2-1/3)^2-(1/3+2)^2]^-1=

$$[ ( 2 - \frac { 1 } { 3 } ) ^ { 2 } - ( \frac { 1 } { 3 } + 2 ) ^ { 2 } ] ^ { - 1 } =$$

Evaluate

$$-\frac{3}{8}=-0.375$$

Solution Steps

Subtract \frac{1}{3} from 2 to get \frac{5}{3}.

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

Calculate \frac{5}{3} to the power of 2 and get \frac{25}{9}.

$$\left(\frac{25}{9}-\left(\frac{1}{3}+2\right)^{2}\right)^{-1}$$

Add \frac{1}{3} and 2 to get \frac{7}{3}.

$$\left(\frac{25}{9}-\left(\frac{7}{3}\right)^{2}\right)^{-1}$$

Calculate \frac{7}{3} to the power of 2 and get \frac{49}{9}.

$$\left(\frac{25}{9}-\frac{49}{9}\right)^{-1}$$

Subtract \frac{49}{9} from \frac{25}{9} to get -\frac{8}{3}.

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

Calculate -\frac{8}{3} to the power of -1 and get -\frac{3}{8}.

$$-\frac{3}{8}$$

Factor

$$-\frac{3}{8} = -0.375$$