Question

If $3x-\frac{y}{5}=10\mathrm{and}xy=5$, then the value of $27x^3-\frac{y^3}{125}$ is ____________.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ 3x-\frac{y}{5}=10...\left(1\right) \\ xy=5...\left(2\right) \\ \mathrm{Now}, \\ 3x-\frac{y}{5}=10 \\ \mathrm{Taking}\mathrm{cube}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(3x-\frac{y}{5}\right)}^3={10}^3 \\ \Rightarrow {\left(3x\right)}^3-{\left(\frac{y}{5}\right)}^3-3\left(3x\right)\left(\frac{y}{5}\right)\left(3x-\frac{y}{5}\right)=1000\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)\right) \\ \Rightarrow 27x^3-\frac{y^3}{125}-\frac{9}{5}xy\left(3x-\frac{y}{5}\right)=1000 \\ \Rightarrow 27x^3-\frac{y^3}{125}-\frac{9}{5}\times 5\times 10=1000\left(\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right) \\ \Rightarrow 27x^3-\frac{y^3}{125}-90=1000 \\ \Rightarrow 27x^3-\frac{y^3}{125}=1000+90 \\ \Rightarrow 27x^3-\frac{y^3}{125}=1090 \end{gathered}$$


Hence, the value of $27x^3-\frac{y^3}{125}$ is 1090.