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

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
$$\begin{vmatrix}
1 & 1-6 & 1-22\\
1-6& 1-14 & 1+82\\ 
1-22 & 1+82 & 1+234\end{vmatrix}=7344$$
