If sin theta + cos theta = p and sin^3 theta | vedclass

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(D) Given,$\sin 	heta + \cos 	heta = p$  $(i)$ and $\sin^3 	heta + \cos^3 	heta = q$  $(ii)$ We know that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. S

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$-2q$

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(D) Given,$\sin 	heta + \cos 	heta = p$  $(i)$ and $\sin^3 	heta + \cos^3 	heta = q$  $(ii)$ We know that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. So,$(\sin 	heta + \cos 	heta)(\sin^2 	heta - \sin 	heta \cos 	heta + \cos^2 	heta) = q$ $p(1 - \sin 	heta \cos 	heta) = q$  (since $\sin^2 	heta + \cos^2 	heta = 1$) $1 - \sin 	heta \cos 	heta = \frac{q}{p}$ $\sin 	heta \cos 	heta = 1 - \frac{q}{p} = \frac{p - q}{p}$  $(iii)$ Now,square equation $(i)$: $(\sin 	heta + \cos 	heta)^2 = p^2$ $\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = p^2$ $1 + 2 \sin 	heta \cos 	heta = p^2$ Substitute $\sin 	heta \cos 	heta$ from $(iii)$: $1 + 2 \left( \frac{p - q}{p} ight) = p^2$ $p + 2p - 2q = p^3$ $3p - 2q = p^3$ $p^3 - 3p = -2q$ $p(p^2 - 3) = -2q$

If $\sin \theta + \cos \theta = p$ and $\sin^3 \theta + \cos^3 \theta = q$,then $p(p^2 - 3)$ is equal to

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