If x^{3}+y^{3}-3 a x y=0,then frac{d y}{d x} | vedclass

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Question

(A) Given equation is $x^{3}+y^{3}-3 a x y=0$. On differentiating both sides with respect to $x$,we get: $\frac{d}{d x}(x^{3}) + \frac{d}{d x}(y^{3})

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$\frac{a y-x^{2}}{y^{2}-a x}$

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(A) Given equation is $x^{3}+y^{3}-3 a x y=0$. On differentiating both sides with respect to $x$,we get: $\frac{d}{d x}(x^{3}) + \frac{d}{d x}(y^{3}) - 3 a \frac{d}{d x}(x y) = 0$ $3 x^{2} + 3 y^{2} \frac{d y}{d x} - 3 a \left( x \frac{d y}{d x} + y \right) = 0$ Dividing by $3$,we get: $x^{2} + y^{2} \frac{d y}{d x} - a x \frac{d y}{d x} - a y = 0$ Grouping the terms containing $\frac{d y}{d x}$: $\frac{d y}{d x} (y^{2} - a x) = a y - x^{2}$ Therefore,$\frac{d y}{d x} = \frac{a y - x^{2}}{y^{2} - a x}$.

If $x^{3}+y^{3}-3 a x y=0$,then $\frac{d y}{d x}$ equals

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