Xavier Aptitude Test 2013 Paper

(e)${\left(\mathit{x}\mathrm{}\mathit{x}\right)}_{\mathrm{}\mathit{b}}=\mathrm{}{\mathit{x}}^3$
In any base system, we can write above equation inthis manner :–
$\mathit{x}\times {\mathit{b}}^2+\mathit{x}\times {\mathit{b}}^1+\mathit{x}\times {\mathit{b}}^0={\mathit{x}}^3$
or $\mathit{x}{\mathit{b}}^2+\mathit{x}\mathit{b}+\mathit{x}-{\mathit{x}}^3=0$
or $\mathit{x}\mathrm{}\left({\mathit{b}}^2+\mathit{b}+1-{\mathit{x}}^2\mathrm{}\right)=0$
or ${\mathit{b}}^2+\mathit{b}+1={\mathit{x}}^2$
$\therefore {\mathit{b}}^2+\mathit{b}+1$ should be a perfect square.
Now check the answer options one by one.
(a) $\mathit{b}=5\Rightarrow {\mathit{b}}^2+\mathit{b}+1=25+5+1=31\ne$ perfect square
(b) $\mathit{b}=6\Rightarrow {\mathit{b}}^2+\mathit{b}+1=36+6+1=43\ne$ perfect square
(c) $\mathit{b}=7\Rightarrow {\mathit{b}}^2+\mathit{b}+1=49+7+1=57\ne$ perfect square
(d) $\mathit{b}=8\Rightarrow {\mathit{b}}^2+\mathit{b}+1=64+8+1=73\ne$ perfect square
So, the only option left is option ( e).