If $\alpha$ and $\beta$ are the zeroes of $p(x) =x^2- px +q = 0$ Find $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.

For convenience, we reverse the sign of $q$, without loss of generality.
$$x^2-px-q$$ is the characteristic equation of the recurrence
$$x_{n+2}=px_{n+1}+qx_n,$$
that has the general solution
$$a\alpha^n+b\beta^n.$$
With $a=b=1$,
$$x_0=2,\\
x_1=\alpha+\beta=p,\\
x_2=\alpha^2+\beta^2=px_1+qx_0=p^2+2q,\\
x_3=\alpha^3+\beta^3=px_2+qx_1=p(p^2+2q)+pq=p^3+3pq,\\
x_4=\alpha^4+\beta^4=px_3+qx_2=p(p^3+3pq)+(p^2+2q)q=p^3+4p^2q+2q^2,\\
x_5=\alpha^5+\beta^5=px_4+qx_3=p(p^3+4p^2q+2q^2)+(p^3+3pq)q=p^4+5p^3q+5pq^2,\\\cdots
$$
The general formula is closely related to the development of $(p+q)^n$.
