Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+3x^2+3x+3=0$, then the value of ${\left(\frac{\alpha }{\alpha +1}\right)}^3+{\left(\frac{\beta }{\beta +1}\right)}^3+{\left(\frac{\gamma }{\gamma +1}\right)}^3$ is

• A. $44$
• B. $45$
• C. $14$
• D. $15$

Answer 1

The correct option is A $44$

Given $\alpha ,\beta ,\gamma$ are the roots of $x^3+2x^2+3x+3=0$ then we have to form the equation whose roots are
$\frac{\alpha }{\alpha +1},\frac{\beta }{\beta +1},\frac{\gamma }{\gamma +1}$
Let, $y=\frac{\alpha }{\alpha +1},\frac{x}{x+1}$ express $x$ in terms of $y$
$\Rightarrow x=\frac{y}{y-1}$
So that the equation is
${\left(\frac{y}{1-y}\right)}^3+2{\left(\frac{y}{1-y}\right)}^2+3\left(\frac{y}{1-y}\right)+3=0$
$\Rightarrow y^3-5y^2+6y-3=0$ ... (i)
Let, $\frac{\alpha }{\alpha +1}={\alpha }^{\prime },\frac{\beta }{\beta }+1={\beta }^{\prime },\frac{\gamma }{\gamma +1}={\gamma }^{\prime }$
Thus ${\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime }$ are the roots of (i)
$\therefore \sum {\alpha }^{\prime }=5,\sum {\alpha }^{\prime }{\beta }^{\prime }=6,\sum {\alpha }^{\prime }{\beta }^{\prime }{\gamma }^{\prime }={\alpha }^{\prime }{\beta }^{\prime }{\gamma }^{\prime }=3$
Now use the identity
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=(a+b+c)\left\{\right(a+b+c{)}^2-3(ab+bc+ca)\}$
Thus
${\alpha }^{\prime 3}+{\beta }^{\prime 3}+{\gamma }^{\prime 3}=({\alpha }^{\prime }+{\beta }^{\prime }+{\gamma }^{\prime })\{{({\alpha }^{\prime }+{\beta }^{\prime }+{\gamma }^{\prime })}^2-3({\alpha }^{\prime }{\beta }^{\prime }+{\beta }^{\prime }{\gamma }^{\prime }+{\gamma }^{\prime }{\alpha }^{\prime }\left)\right\}+3{\alpha }^{\prime }{\beta }^{\prime }{\gamma }^{\prime }$
$=\left(5\right)\left[\right(5{)}^2-3\left(6\right)]+3(3)$
$=5[25-18]+9$
$=35+9=44$