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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) $-\frac{1}{2}$

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Solution

## $\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\frac{x}{y}+\frac{y}{x}=1\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}+{y}^{2}}{xy}=1\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=xy\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-xy=0...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\left({x}^{3}+{y}^{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x+y\right)\left({x}^{2}+{y}^{2}-xy\right)\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)\right)\phantom{\rule{0ex}{0ex}}=\left(x+y\right)×0\left(\mathrm{From}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{c}\right).$

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