Since α and β are roots of equation x2+px+q=0,50 α+β=−p and αβ=q As we know that, α3+β3=(α+β)(α2+β2−αβ) =(α+β)[(α+β)2−3αβ] =(−p)[p2−3q]=3pq−p3 and α4+α2β2+β4 =(α2+β2)2−α2β2=[(α+β)2−2αβ]2−(αβ)2 =((−p)2−2q)2−q2 =(p2−2q)2−q2=(p2−2q−q)(p2−2q+q) =(p2−q)(p2−3q)