<s>[INST] 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
 [/INST] 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.
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