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). Using the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

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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). Using the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,we have: $(\sin 	heta + \cos 	heta)(\sin^2 	heta - \sin 	heta \cos 	heta + \cos^2 	heta) = q$. Since $\sin^2 	heta + \cos^2 	heta = 1$,this becomes $p(1 - \sin 	heta \cos 	heta) = q$. Thus,$1 - \sin 	heta \cos 	heta = \frac{q}{p}$,which implies $\sin 	heta \cos 	heta = 1 - \frac{q}{p}$ (iii). Squaring equation $(i)$: $(\sin 	heta + \cos 	heta)^2 = p^2$. $\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = p^2$. $1 + 2 \sin 	heta \cos 	heta = p^2$. Substituting (iii) into this: $1 + 2(1 - \frac{q}{p}) = p^2$. $1 + 2 - \frac{2q}{p} = p^2$. $3 - \frac{2q}{p} = p^2$. Multiply by $p$: $3p - 2q = p^3$. Rearranging gives $p^3 - 3p = -2q$,or $p(p^2 - 3) = -2q$.

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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