Question

If $x=2+\sqrt{3}$, find : $x^3+\frac{1}{x^3}$.

Answer 1

$x=2+\sqrt{3}$

Taking cube of both sides, we get

$x^3={(2+\sqrt{3})}^3$

$x^3=2^3+(\sqrt{3}{)}^3+3\times 2\times \sqrt{3}(2+\sqrt{3}\left)\right[(a+b{)}^3=a^3+b^3+3ab(a+b\left)\right]$

$x^3=8+3\sqrt{3}+6\sqrt{3}(2+\sqrt{3})$

$x^3=8+3\sqrt{3}+12\sqrt{3}+18$

$x^3=26+15\sqrt{3}$

$x^3+\frac{1}{x^3}=(26+15\sqrt{3})+\left(\frac{1}{26+15\sqrt{3}}\right)$

$=\frac{(26+15\sqrt{3}{)}^2+1}{(26+15\sqrt{3})}$

$=\frac{(26{)}^2+(15\sqrt{3}{)}^2+2\times 26\times 15\sqrt{3}}{(26+15\sqrt{3})}\left[\right(a+b{)}^2=a^2+b^2+2ab]$

$=\frac{676+675+780\sqrt{3}}{(26+15\sqrt{3})}$

$=\frac{1351+780\sqrt{3}}{(26+15\sqrt{3})}$