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If $x^{3}+y^{3}-3 a x y=0$, then $\frac{d y}{d x}$ equals
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Verified Answer
The correct answer is:
$\frac{a y-x^{2}}{y^{2}-a x}$
Given, $x^{3}+y^{3}-3 a x y=0$
On differentiating w.r.t. $x$, we get $3 x^{2}+3 y^{2} \cdot \frac{d y}{d x}-3 a\left(x \frac{d y}{d x}+y\right)=0$
$\Rightarrow 3\left(x^{2}-a y\right)+3 \frac{d y}{d x}\left(y^{2}-a x\right)=0$
$\Rightarrow$
$\frac{d y}{d x}=\frac{a y-x^{2}}{y^{2}-a x}$
On differentiating w.r.t. $x$, we get $3 x^{2}+3 y^{2} \cdot \frac{d y}{d x}-3 a\left(x \frac{d y}{d x}+y\right)=0$
$\Rightarrow 3\left(x^{2}-a y\right)+3 \frac{d y}{d x}\left(y^{2}-a x\right)=0$
$\Rightarrow$
$\frac{d y}{d x}=\frac{a y-x^{2}}{y^{2}-a x}$
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