Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-6x+6=0$, what is ${\alpha }^3+{\beta }^3+{\alpha }^2+{\beta }^2+\alpha +\beta$ equal to?

• A. $150$
• B. $138$
• C. $128$
• D. $124$

Answer 1

The correct option is B $138$

Given quadratic equation is $x^2-6x+6=0$
$\alpha +\beta =6,\alpha \beta =6$
We know ${\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$
$=6^3-3\times 6\times 6$
$=216-108=108$
Also ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =6^2-2\times 6=36-12=24$
Thus the value of $({\alpha }^3+{\beta }^3)+({\alpha }^2+{\beta }^2)+(\alpha +\beta )$ is $108+24+6=138$.