SSC CHSL 17 Mar 2020 Shift 3 Paper

Let the two numbers be $\mathit{x}$ and $\mathit{y}$.
Then, according to question,
$\mathit{x}-\mathit{y}=6$ ...(i)
${\mathit{x}}^2-{\mathit{y}}^2=60$
⇒ $\left(\mathit{x}+\mathit{y}\right)\left(\mathit{x}-\mathit{y}\right)=60$
⇒ $\mathit{x}+\mathit{y}=\frac{60}{6}=10$ [from Eq. (i)]
⇒ $\mathit{x}+\mathit{y}=10$ ...(ii)
On, solving Eqs. (i) and (ii), we get
$\mathit{x}=8,\mathit{y}=2$
Now, sum of their cubes $={\mathit{x}}^3+{\mathit{y}}^3$
$=8^3+2^3$
$=512+8$
$=520$