If $\alpha,\ \beta$ are the roots of $x^2-px+q=0$, find the value of: $( i). \alpha^2+\beta^2\ ( ii). \alpha^3+\beta^3$.

Given: $\alpha,\ \beta$  are the roots of $x^2-px+q=0$.

To do: To find the value of: $( i). \alpha^2+\beta^2\ ( ii). \alpha^3+\beta^3$.

Solution:

As given, $\alpha,\ \beta$  are the roots of $x^2-px+q=0$.

On comparing the given equation to $ax^2+bx+c=0$,

Here, $a=1,\ b=-p,\ c=q$

Therefore, sum of the roots$=\alpha+\beta=-\frac{-p}{1}=p$

Product of the roots$=\alpha\beta=\frac{c}{a}=\frac{q}{1}=q$

$( i). \alpha^2+\beta^2$

$=( \alpha+\beta)^2-2\alpha\beta$

$=p^2-2q$                                  [$\because \alpha+\beta=p,\ \alpha\beta=q$]

Thus, $\alpha^2+\beta^2=p^2-2q$.

$( ii). \alpha^3+\beta^3$

$=( \alpha+\beta)^3-3\alpha\beta( \alpha+\beta)$

$=p^3-3pq$                                           [$\because \alpha+\beta=p,\ \alpha\beta=q$]

Thus, $\alpha^3+\beta^3=p^3-3pq$