Given, sinθ+cosθ=p ...(i)
and sin3θ+cos3θ=q ...(ii) ⇒(sinθ+cosθ) (sin2θ−sinθ⋅cosθ+cos2θ)=q ⇒p(1−sinθ⋅cosθ)=q
[From Eq. (i) and sin2θ+cos2θ=1] ⇒1−sinθ⋅cosθ=pq ⇒sinθ⋅cosθ=1−pq ...(iii)
On squaring both sides of Eq. (i), we get sin2θ+cos2θ+2sinθ⋅cosθ=p2 ⇒1+2(1−pq)=p2 [from Eq. (iii)] ⇒p+2(p−q)=p3 ⇒3p−2q=p3 ⇒p3−3p=−2q ⇒p(p2−3)=−2q