SSC CHSL 6 Dec 2015 Shift 2 Paper

Given : ${\left(2\mathit{a}-1\right)}^2+{\left(4\mathit{b}-3\right)}^2$ $+{\left(4\mathit{c}+5\right)}^2=0$
∵ Sum of 3 positive terms is 0, then each term is equal to '0'.
⇒ $2\mathit{a}-1=0$
⇒ $\mathit{a}=\frac{1}{2}=\frac{2}{4}$
Similarly, $\mathit{b}=\frac{3}{4}$ and $\mathit{c}=-\frac{5}{4}$
Now, $\left(\mathit{a}+\mathit{b}+\mathit{c}\right)=\frac{2}{4}+\frac{3}{4}$ $+\left(-\frac{5}{4}\right)=0$ ----(i)
Using,${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3-3\mathit{a}\mathit{b}\mathit{c}$ $=\left(\mathit{a}+\mathit{b}+\mathit{c}\right)$ $\left({\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2-\mathit{a}\mathit{b}-\mathit{b}\mathit{c}-\mathit{a}\mathit{c}\right)$
⇒ ${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3-3\mathit{a}\mathit{b}\mathit{c}$ $\left(0\right)\left({\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2-\mathit{a}\mathit{b}-\mathit{b}\mathit{c}-\mathit{a}\mathit{c}\right)$ [Using equation (i)]
⇒ ${\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3-3\mathit{a}\mathit{b}\mathit{c}=0$ --------(ii)
To find : $\frac{{\mathit{a}}^3+{\mathit{b}}^3+{\mathit{c}}^3-3\mathit{a}\mathit{b}\mathit{c}}{{\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2}$
= $\frac{0}{{\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2}$ = 0 [Using equation (ii)]