Find the Determinant when $p$ and $q$ are roots of $x^2-2x+5=0$

Question

If $p$ and $q$ are roots of $x^2-2x+5=0$ Then find value of

$$\Delta=\begin{vmatrix} 1 & 1+p^2+q^2 & 1+p^3+q^3\\ 1+p^2+q^2& 1+p^4+q^4 & 1+p^5+q^5\\ 1+p^3+q^3 & 1+p^5+q^5 & 1+p^6+q^6 \end{vmatrix}$$

My try: I tried to express the determinant as product of two determinants but in vain.

Secondly i tried taking $$p=\sqrt{5}\frac{1+2i}{\sqrt{5}}=\sqrt{5}(\cos t+i\sin t)$$ and $q$ its conjugate, but still its lengthy

Answer 1

Use Viète's formulas: $$p+q=2\qquad pq=5$$ $$p^2+q^2=(p+q)^2-2pq=-6$$ $$p^3+q^3=(p+q)(p^2+q^2-pq)=-22$$ $$p^4+q^4=(p^2+q^2)^2-2(pq)^2=-14$$ $$p^5+q^5=(p+q)(p^4+q^4)-pq(p^3+q^3)=82$$ $$p^6+q^6=(p^3+q^3)^2-2(pq)^3=234$$ from which we may easily work out the determinant as $$\begin{vmatrix} 1 & 1-6 & 1-22\\ 1-6& 1-14 & 1+82\\ 1-22 & 1+82 & 1+234\end{vmatrix}=7344$$

Answer 2

HINT:

Note that by Vieta's formulas, $p+q=2$ and $pq=5$ so $$1+p+q=3$$ and $$1+p^2+q^2=1+2p-5+2q-5=2(p+q)-9=-5$$ giving $$1+p^3+q^3=(1+p+q)(1+p^2+q^2)-(p^2+q^2+p+q+pq(p+q))$$ and since $x^2=2x-5$, $$1+p^3+q^3=3(-5)-(2p-5+2q-5+2+5(2))$$ or $$1+p^3+q^3=-15-(2(p+q)+2)=-15-(6)=-21$$ and so on.