If $\alpha$ and $\beta$ are the zeroes of $p(x) =x^2- px +q = 0$ Find $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.
 A: We have, $$x^2-px+q=0 $$ $$\implies \alpha+\beta=\frac{-(-p)}{1}=p$$ & $$\alpha\beta=\frac{q}{1}=q$$ Now, we have $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$
$$=(p)^2-2q=p^2-2q$$ 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\alpha^2+\beta^2=p^2-2q}}$$
& $$\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta)$$
$$=(p)(p^2-2q-q)=p^3-3pq$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\alpha^3+\beta^3=p^3-3pq}}$$
A: $\bf{My\; Solution::}$ Given $\alpha$ and $\beta$ are two roots of $x^2-px+q=0$
So $$\displaystyle \alpha+\beta = p$$ and $$\alpha \cdot \beta = q$$
Now $$\alpha^3+\beta^3 = \left(\alpha+\beta\right)^3-3\alpha \cdot \beta \left(\alpha+\beta\right) = p^3-3pq$$
and $$\alpha^2+\beta^2 = \left(\alpha+\beta\right)^2-2\alpha \cdot \beta = p^2-2q$$
A: 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 $(p+q)^n$.
A: You have $p=\alpha+\beta$ and $q=\alpha \beta$
Now $p(\alpha)+p(\beta)=0$
Expanding the two expressions gives $\alpha^2+\beta^2-p(\alpha+\beta)+2q=\alpha^2+\beta^2-p^2+2q=0$
Note also that $\alpha p(\alpha)+\beta p(\beta) =0$ and this gives $$(\alpha^3+\beta^3)-p(\alpha^2+\beta^2)+q(\alpha+\beta)=0$$And this can be used to compute $\alpha^3+\beta^3$ from what is already known.
With a little thought this can be generalised to a range of other situations.
