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