If sqrt{1-x^6}+sqrt{1-y^6}=a(x^3-y^3),then y^ | vedclass

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(C) Let $x^3 = \sin 	heta$ and $y^3 = \sin \phi$. Then the equation becomes $\cos 	heta + \cos \phi = a(\sin 	heta - \sin \phi)$.Using trigonometri

Vedclass · Accepted Answer

$x^2 \sqrt{\frac{1-y^6}{1-x^6}}$

Vedclass · Answer

(C) Let $x^3 = \sin 	heta$ and $y^3 = \sin \phi$. Then the equation becomes $\cos 	heta + \cos \phi = a(\sin 	heta - \sin \phi)$.Using trigonometric identities,$2 \cos \frac{	heta+\phi}{2} \cos \frac{	heta-\phi}{2} = a \cdot 2 \cos \frac{	heta+\phi}{2} \sin \frac{	heta-\phi}{2}$.This implies $\cot \frac{	heta-\phi}{2} = a$,which is a constant.Thus,$\frac{	heta-\phi}{2} = 	ext{constant} \Rightarrow 	heta - \phi = C$.Substituting back,$\arcsin(x^3) - \arcsin(y^3) = C$.Differentiating with respect to $x$:$\frac{3x^2}{\sqrt{1-x^6}} - \frac{3y^2}{\sqrt{1-y^6}} \frac{dy}{dx} = 0$.Rearranging the terms,we get $\frac{3y^2}{\sqrt{1-y^6}} \frac{dy}{dx} = \frac{3x^2}{\sqrt{1-x^6}}$.Therefore,$y^2 \frac{dy}{dx} = x^2 \sqrt{\frac{1-y^6}{1-x^6}}$.

If $\sqrt{1-x^6}+\sqrt{1-y^6}=a(x^3-y^3)$,then $y^2 \frac{dy}{dx}=$

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