Question

if x²+y²=29 and xy=10, then find the value of x³-y³

Answer 1

$$\begin{gathered} x^2+y^2=29 \\ \Rightarrow {\left(x-y\right)}^2+2xy=29 \\ \Rightarrow {\left(x-y\right)}^2+2\times 10=29 \\ \Rightarrow {\left(x-y\right)}^2=29-20 \\ \Rightarrow {\left(x-y\right)}^2=9 \\ \Rightarrow \left(x-y\right)=\sqrt{9}=3...\left(\mathrm{i}\right) \\ \Rightarrow {\left(x-y\right)}^3=3^3\{\mathrm{Cubing}\mathrm{both}\mathrm{sides}\} \\ \Rightarrow x^3-y^3-3xy\left(x-y\right)=27\{\mathrm{using}\mathrm{the}\mathrm{identity}{\left(\mathrm{a}-\mathrm{b}\right)}^3={\mathrm{a}}^3-{\mathrm{b}}^3-3\mathrm{ab}\left(\mathrm{a}-\mathrm{b}\right)\} \\ \Rightarrow x^3-y^3-3\times 10\times 3=27\{\mathrm{using}(\mathrm{i})\mathrm{and}\mathrm{given}xy=10\} \\ \Rightarrow x^3-y^3-90=27 \\ \Rightarrow x^3-y^3=117 \end{gathered}$$