# Irrational numbers

• @Eliza-Berthman

Notice that the denominator is of the form of $a^{\frac{1}{3}}-b^{\frac{1}{3}}$ [a=2,b=1]

We know $a^3-b^3=(a-b)(a^2+b^2+ab)$
We can also write this as,
$a-b=(a^{\frac{1}{3}}-b^{\frac{1}{3}})(a^{\frac{2}{3}}+b^{\frac{2}{3}}+a^{\frac{1}{3}}\cdot b^{\frac{1}{3}})$

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).

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