Question

If ${\alpha }^3-5{\alpha }^2+5\alpha -3=0$
${\beta }^3-5{\beta }^2+5\beta -3=0$
${\gamma }^3-5{\gamma }^2+5\gamma -3=0$.
then find $\alpha +\beta +\gamma$ is
($\alpha ,\beta ,\gamma$ are distinct)

• A. $5$
• B. $-5$
• C. $3$
• D. $1$

Answer 1

The correct option is A $5$

${\alpha }^3-5{\alpha }^2+5\alpha -3=0$

${\beta }^3-5{\beta }^2+5\beta -3=0$

${\gamma }^3-5{\gamma }^2+5\gamma -3=0$

From

these equations we can say that $\alpha ,\beta ,\gamma$

are the roots of the equation

$x^3-5x^2+5x-3=0$

$\alpha +\beta +\gamma =-b/a=-(-5)/1=5$

$\therefore \alpha +\beta +\gamma =5$