Question

If $a=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}},b=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, find $a^3+b^3$.

Answer 1

Let us first rationalize the denominators of the given $a$ and $b$ as follows:

$a=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2}{)}^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\frac{(\sqrt{3}{)}^2+(\sqrt{2}{)}^2+(2\times \sqrt{3}\times \sqrt{2})}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}(\because (a-b\left)\right(a+b)=a^2-b^2,(a+b{)}^2=a^2+b^2+2ab)=\frac{3+2+2\sqrt{6}}{3-2}$

$=\frac{5+2\sqrt{6}}{1}\Rightarrow a=5+2\sqrt{6}$

$b=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2}{)}^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{(\sqrt{3}{)}^2+(\sqrt{2}{)}^2-(2\times \sqrt{3}\times \sqrt{2})}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}(\because (a-b\left)\right(a+b)=a^2-b^2,(a-b{)}^2=a^2+b^2-2ab)=\frac{3+2-2\sqrt{6}}{3-2}$

$=\frac{5-2\sqrt{6}}{1}\Rightarrow b=5-2\sqrt{6}$

We know that one of the identity of polynomials is:

$a^3+b^3=(a+b)(a^2+b^2-ab)$

Substituting the values of $a$ and $b$ we get,

$a^3+b^3=(a+b)(a^2+b^2-ab)=(5+2\sqrt{6}+5-2\sqrt{6})\left[\right(5+2\sqrt{6}{)}^2+(5-2\sqrt{6}{)}^2-(5+2\sqrt{6}\left)\right(5-2\sqrt{6}\left)\right]=10\left[\right(5^2+(2\sqrt{6}{)}^2+(2\times 5\times 2\sqrt{6}\left)\right)+(5^2+(2\sqrt{6}{)}^2-(2\times 5\times 2\sqrt{6}))-(5^2-\left(2\sqrt{6}{)}^2\right)](\because (a+b{)}^2=a^2+b^2+2ab,(a-b{)}^2=a^2+b^2-2ab,(a+b\left)\right(a-b)=a^2-b^2)=10[25+24+20\sqrt{6}+25+24-20\sqrt{6}-(25-24\left)\right]=10(98-1)=10\times 97=970$

Hence, $a^3+b^3=970$.