Question

If $\frac{x}{y}+\frac{y}{x}=1\left(x,y\ne 0\right)$, the value of x³ + y³ is
(a) 1
(b) –1
(c) 0
(d) $-\frac{1}{2}$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \frac{x}{y}+\frac{y}{x}=1 \\ \Rightarrow \frac{x^2+y^2}{xy}=1 \\ \Rightarrow x^2+y^2=xy \\ \Rightarrow x^2+y^2-xy=0...\left(1\right) \\ \mathrm{Now}, \\ \left(x^3+y^3\right) \\ =\left(x+y\right)\left(x^2+y^2-xy\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\right) \\ =\left(x+y\right)\times 0\left(\mathrm{From}\left(1\right)\right) \\ =0 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{c}\right). \end{gathered}$$