1. If xy(x+y)=1 then, the value of $$\frac{1}{x^{3}y^{3}}-x^{3}-y^{3}$$ is
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By: anil on 05 May 2019 02.58 pm
xy(x+y)=1 x+y = 1/xy apply cube on both sides, $$(x+y)^{3}$$ = $$frac{1}{x^{3}y^{3}}$$ $$x^{3}+y^{3}+3xy(x+y)$$ = $$frac{1}{x^{3}y^{3}}$$
$$x^{3}+y^{3}+3(1)$$ = $$frac{1}{x^{3}y^{3}}$$
3 = $$frac{1}{x^{3}y^{3}}$$ - $$x^{3}-y^{3}$$ so the answer is option A.
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$$x^{3}+y^{3}+3(1)$$ = $$frac{1}{x^{3}y^{3}}$$
3 = $$frac{1}{x^{3}y^{3}}$$ - $$x^{3}-y^{3}$$ so the answer is option A.