Question

If $2x+\frac{y}{3}=12\mathrm{and}xy=30,\mathrm{then}8x^3+\frac{y^3}{27}=\_\_\_\_\_\_\_$
(a) 1008
(b) 168
(c) 106
(d) none of these

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ 2x+\frac{y}{3}=12...\left(1\right) \\ xy=30...\left(2\right) \\ \mathrm{Now}, \\ 2x+\frac{y}{3}=12 \\ \mathrm{Taking}\mathrm{cube}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(2x+\frac{y}{3}\right)}^3={12}^3 \\ \Rightarrow {\left(2x\right)}^3+{\left(\frac{y}{3}\right)}^3+3\left(2x\right)\left(\frac{y}{3}\right)\left(2x+\frac{y}{3}\right)=1728\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right) \\ \Rightarrow 8x^3+\frac{y^3}{27}+2xy\left(2x+\frac{y}{3}\right)=1728 \\ \Rightarrow 8x^3+\frac{y^3}{27}+2\times 30\times 12=1728\left(\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right) \\ \Rightarrow 8x^3+\frac{y^3}{27}+720=1728 \\ \Rightarrow 8x^3+\frac{y^3}{27}=1728-720 \\ \Rightarrow 8x^3+\frac{y^3}{27}=1008 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{a}\right). \end{gathered}$$