Irrational numbers


Notice that the denominator is of the form of
$a^{\frac{1}{3}}b^{\frac{1}{3}}$ [a=2,b=1]We know
$a^3b^3=(ab)(a^2+b^2+ab)$
We can also write this as,
$ab=(a^{\frac{1}{3}}b^{\frac{1}{3}})(a^{\frac{2}{3}}+b^{\frac{2}{3}}+a^{\frac{1}{3}}\cdot b^{\frac{1}{3}})$ Observing this we multiply the numerator and denominator by
$(2^{\frac{2}{3}}+1^2+2^{\frac{1}{3}}\cdot1)$ and using above identity we get 1 in the denominator and$4^{\frac{1}{3}}+2^{\frac{1}{3}}+1$ in the numerator,i.e., option c).