(c) We have $x\,{\sin ^3}\alpha + y\,{\cos ^3}\alpha = \sin \,\alpha \,\cos \,\alpha $…..$(i)$

and $x\,\sin \,\alpha – y\,\cos \,\alpha = 0$…..$(ii)$

Now from $(ii)$, $x\,\sin \,\alpha = y\,\cos \,\alpha $

Putting in $(i),$ we get

$ \Rightarrow \,\,y\,\cos \alpha \,{\sin ^2}\alpha + y\,{\cos ^3}\alpha = \sin \,\alpha \,\cos \,\alpha $

$ \Rightarrow \,\,y\,\cos \alpha \,\left\{ {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right\} = \sin \,\alpha \,\cos \,\alpha $

$ \Rightarrow \,\,y\,\cos \,\alpha = \sin \,\alpha \,\cos \,\alpha \, $

$\Rightarrow \,\,y = \sin \,\alpha $ and $x = \cos \,\alpha $

Hence, ${x^2} + {y^2} = {\sin ^2}\alpha + {\cos ^2}\alpha = 1.$