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Mathematics Question on Complex Numbers and Quadratic Equations

If α,β\alpha, \beta are the roots of the quadratic equation x2+px+q=0x^2 + px + q = 0, then the values of α3,β3\alpha^{3}, \beta^{3} and α4+α2β3+β4\alpha^{4}+\alpha^{2}\beta^{3}+\beta^{4} are respectively

A

3pq−p33pq - p^3 and p4−3p2q+3q2p^4 - 3p^2q + 3q^2

B

−p(3q−p2)-p(3q - p^2) and (p2−q)(p2+3q)(p^2 - q)(p^2 + 3q)

C

pq−4pq - 4 and p4−q4p^4 - q^4

D

3pq−p33pq - p^3 and (p2−q)(p2−3q)(p^2 - q) (p^2 - 3q)

Answer

3pq−p33pq - p^3 and (p2−q)(p2−3q)(p^2 - q) (p^2 - 3q)

Explanation

Solution

The correct answer is D:3pq−p3  and  (p2−q)(p2−3q)3pq-p^3\space and \space (p^2-q)(p^2-3q)
Given that;
Quadratic equation is x2+px+q=0x^2+px+q=0
∵Sum  of  roots,α+β=−pandαβ=q\because Sum\space of\space roots, \alpha+\beta=-p and \alpha \beta=q
∴(α3+β3)=(α+β)3−3αβ(α+β)\therefore\left(\alpha^{3}+\beta^{3}\right)=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)
=(−p)3−3q(−p)=(-p)^{3}-3 q(-p)
=−p3+3pq=-p^{3}+3 p q
and  α4+α2β2+β4=(α4+β4)+(αβ)2and\space \alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4}=\left(\alpha^{4}+\beta^{4}\right)+(\alpha \beta)^{2}
=(α2+β2)2−(αβ)2=\left(\alpha^{2}+\beta^{2}\right)^{2}-(\alpha \beta)^{2}
=[(α+β)2−2αβ]2−(αβ)2= {\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]^{2}-(\alpha \beta)^{2}}
=[(−p)2−2q]2−3(q)2=\left[(-p)^{2}-2 q\right]^{2}-3(q)^{2}
=[p2−2q]2−3q2=\left[p^{2}-2 q\right]^{2}-3 q^{2}
=p4−4p2q+4q2−q2=p^{4}-4 p^{2} q+4 q^{2}-q^{2}
=p4−4p2q+3q2=p^{4}-4 p^{2} q+3 q^{2}
=p4−3p2q−p2q+3q2=p^{4}-3 p^{2} q-p^{2} q+3 q^{2}
=p2(p2−3q)−q(p2−3q)=p^{2}\left(p^{2}-3 q\right)-q\left(p^{2}-3 q\right)
=(p2−q)(p2−3q)= \left(p^{2}-q\right)\left(p^{2}-3 q\right)
Quadratic equation