SSC CHSL 5 Mar 2018 Shift 1 Paper

Given : ${\mathit{r}}^3+{\mathit{s}}^3=0$ and $\mathit{r}+\mathit{s}=6$ ---------(i)
∵ ${\mathit{x}}^3+{\mathit{y}}^3=\left(\mathit{x}+\mathit{y}\right)\left({\mathit{x}}^2+{\mathit{y}}^2-\mathit{x}\mathit{y}\right)$
⇒ $\left(\mathit{r}+\mathit{s}\right)\left({\mathit{r}}^2+{\mathit{s}}^2-\mathit{r}\mathit{s}\right)=0$
⇒ $6\left({\mathit{r}}^2+{\mathit{s}}^2-\mathit{r}\mathit{s}\right)=0$
⇒ ${\mathit{r}}^2+{\mathit{s}}^2-\mathit{r}\mathit{s}=0$
⇒ ${\mathit{r}}^2+{\mathit{s}}^2=\mathit{r}\mathit{s}$ ----------(ii)
Also, squaringequation (i),
⇒ ${\left(\mathit{r}+\mathit{s}\right)}^2={\left(6\right)}^2$
⇒ ${\mathit{r}}^2+{\mathit{s}}^2+2\mathit{r}\mathit{s}=36$
Substituting value from equation (ii), ⇒ rs + 2rs = 3rs = 36
⇒ $\mathit{r}\mathit{s}=\frac{36}{3}=12$------------(iii)
∴ $\left(\frac{1}{\mathit{r}}+\frac{1}{\mathit{s}}\right)=\frac{\mathit{r}+\mathit{s}}{\mathit{r}\mathit{s}}$
Dividing equation (i) by (iii),
= $\frac{6}{12}=0.5$