SSC CGL 2 Sep 2016 Shif 2 Paper

a + b + c = 15,
$\frac{1}{\mathit{a}}+\frac{1}{\mathit{b}}+\frac{1}{\mathit{c}}$ = $\frac{71}{\mathit{a}\mathit{b}\mathit{c}}$
⇒ $\frac{\mathit{b}\mathit{c}+\mathit{a}\mathit{c}+\mathit{a}\mathit{b}}{\mathit{a}\mathit{b}\mathit{c}}$ = $\frac{71}{\mathit{a}\mathit{b}\mathit{c}}$
⇒ ab + bc + ca = 71
∴ ${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3$ - 3abc = (a + b + c) (${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2$ - ab - bc - ac)
= (a + b + c) {${\left(\mathit{a}+\mathit{b}+\mathit{c}\right)}^2$ - 3 (ab + bc + ca)}
= 15 (${15}^2$ - 3 × 71)
= 15 (225 - 213) = 15 × 12
= 180