Solve left(2^-1-3^-1right)^-1+{left(6^-1-{8}^-1right)}^-1

$${ \left( { 2 }^{ -1 } - { 3 }^{ -1 } \right) }^{ -1 } + { \left( { 6 }^{ -1 } - { 8 }^{ -1 } \right) }^{ -1 }$$

Evaluate

$$30$$

Solution Steps

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

$$\left(\frac{1}{2}-3^{-1}\right)^{-1}+\left(6^{-1}-8^{-1}\right)^{-1}$$

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

$$\left(\frac{1}{2}-\frac{1}{3}\right)^{-1}+\left(6^{-1}-8^{-1}\right)^{-1}$$

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

$$\left(\frac{1}{6}\right)^{-1}+\left(6^{-1}-8^{-1}\right)^{-1}$$

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

$$6+\left(6^{-1}-8^{-1}\right)^{-1}$$

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

$$6+\left(\frac{1}{6}-8^{-1}\right)^{-1}$$

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

$$6+\left(\frac{1}{6}-\frac{1}{8}\right)^{-1}$$

Subtract \frac{1}{8} from \frac{1}{6} to get \frac{1}{24}.

$$6+\left(\frac{1}{24}\right)^{-1}$$

Calculate \frac{1}{24} to the power of -1 and get 24.

$$6+24$$

Add 6 and 24 to get 30.

$$30$$

Factor

$$2\times 3\times 5$$