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Question Number 65651 by Masumsiddiqui399@gmail.com last updated on 01/Aug/19

Commented by Prithwish sen last updated on 01/Aug/19

(x−1)^2 +y^2 =0 ⇒x−1=0 and y=0  ∴x^3 +y^3 =1

$$\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\mathrm{x}−\mathrm{1}=\mathrm{0}\:\mathrm{and}\:\mathrm{y}=\mathrm{0} \\ $$$$\therefore\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{1} \\ $$

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