Find the value of  $$(\frac{1}{6})^{-3} + (\frac{1}{2})^{-3} + (\frac{1}{3})^{-3}$$​​

Correct option is A

Given: 

$$(\frac{1}{6})^{-3} + (\frac{1}{2})^{-3} + (\frac{1}{3})^{-3}$$​

Concept Used: 

$$\frac{1}{n} = n^{-1}$$​

Solution: 

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

$$(6 ^{-1})^{-3} + (2^{-1})^{-3} + (3^{-1})^{-3}$$

=  $$6^3 + 2^3 +3^3$$ 

= 216 + 8 + 27 

= 251