If x/y + y/x = −1, Find the Value of x³ − y³
If \[ \frac{x}{y}+\frac{y}{x}=-1 \quad (x,y\ne0), \] then the value of \[ x^3-y^3 \] is
(a) \(1\)
(b) \(-1\)
(c) \(0\)
(d) \(\frac{1}{2}\)
Solution
\[ \frac{x}{y}+\frac{y}{x}=-1 \]
Multiplying by \(xy\):
\[ x^2+y^2=-xy \]
\[ x^2+xy+y^2=0 \]
Using identity:
\[ x^3-y^3=(x-y)(x^2+xy+y^2) \]
\[ =(x-y)\times0 \]
\[ =0 \]
Therefore,
\[ \boxed{(c)\ 0} \]