SSC CHSL 12 Oct 2020 Shift 3 Paper

Let the two numbers be $\mathit{x}$ and $\mathit{y}$.
Then, $\mathit{x}+\mathit{y}=11$ ...(i)
and${\mathit{x}}^2+{\mathit{y}}^2=65$ ...(ii)
Now, on squaring Eq. (i) both sides, we get
${\left(\mathit{x}+\mathit{y}\right)}^2=121$
⇒ ${\mathit{x}}^2+{\mathit{y}}^2+2\mathit{x}\mathit{y}=121$
⇒ ${\mathit{x}}^2+{\mathit{y}}^2=121-2\mathit{x}\mathit{y}$
⇒ ${\mathit{x}}^2+{\mathit{y}}^2=121-2\times \mathit{x}\mathit{y}$
⇒ $65=121-2\mathit{x}\mathit{y}$
⇒ $2\mathit{x}\mathit{y}=121-65$
⇒ $2\mathit{x}\mathit{y}=56$
⇒ $\mathit{x}\mathit{y}=28$ ...(iii)
Now, we know that,
${\mathit{x}}^3+{\mathit{y}}^3=\left(\mathit{x}+\mathit{y}\right)\left({\mathit{x}}^2+{\mathit{y}}^2-\mathit{x}\mathit{y}\right)$
From Eqs. (ii) and (iii), we get
⇒ ${\mathit{x}}^3+{\mathit{y}}^3=11\times \left(65-28\right)$
${\mathit{x}}^3+{\mathit{y}}^3=11\times 37$
∴ ${\mathit{x}}^3+{\mathit{y}}^3=407$