Question:
If \[ \frac{x}{y}+\frac{y}{x}=-1 \quad (x,y \ne 0), \] then the value of \[ x^3+y^3 \] is
(a) 1
(b) -1
(c) 0
(d) \(\frac{1}{2}\)
Solution:
\[ \frac{x}{y}+\frac{y}{x}=-1 \]
\[ \frac{x^2+y^2}{xy}=-1 \]
\[ x^2+y^2=-xy \]
\[ x^2+xy+y^2=0 \]
Using identity
\[ x^3+y^3=(x+y)(x^2-xy+y^2) \]
\[ =(x+y)\big[(x^2+y^2)-xy\big] \]
\[ =(x+y)(-xy-xy) \]
\[ =-2xy(x+y) \]
Also,
\[ x^2+xy+y^2=0 \]
\[ (x+y)^2=xy \]
Hence,
\[ x^3+y^3=1 \]
\[ \boxed{1} \]