Question

Let $p,q,r$ be roots of cubic equation $x^3+2x^2+3x+3=0$, then

• A. ${\left(\frac{p}{p+1}\right)}^3+{\left(\frac{q}{q+1}\right)}^3+{\left(\frac{r}{r+1}\right)}^3=44$
• B. $\frac{p}{p+1}+\frac{q}{q+1}+\frac{r}{r+1}=6$
• C. ${\left(\frac{p}{p+1}\right)}^3+{\left(\frac{q}{q+1}\right)}^3+{\left(\frac{r}{r+1}\right)}^3=38$
• D. $\frac{p}{p+1}+\frac{q}{q+1}+\frac{r}{r+1}=5$

Answer 1

Answer: (A), (D)

$\frac{p}{p+1}+\frac{q}{q+1}+\frac{r}{r+1}=5$

Using transformation
$\text{Let}\frac{p}{p+1}=y\Rightarrow p=\frac{y}{1-y}$
$p$ is a root of the given cubic equation.
So $p^3+2p^2+3p+3=0$
$\Rightarrow {\left(\frac{y}{1-y}\right)}^3+2{\left(\frac{y}{1-y}\right)}^2+3\left(\frac{y}{1-y}\right)+3=0$
$\Rightarrow y^3-5y^2+6y-3=0$
So $y_1,y_2,y_3$ are root of above equation, so
$\frac{p}{p+1}+\frac{q}{q+1}+\frac{r}{r+1}=5$
$\sum y_i=5,\sum y_i{y}_j=6,\sum y_i{y}_j{y}_k=3$
$y_1^3+y_2^3+y_3^3-3y_1{y}_2{y}_3=(\sum y_i)\left[\right(\sum y_i{)}^2-3\sum y_i{y}_j]$
$\Rightarrow y_1^3+y_2^3+y_3^3-3\left(3\right)=5(5^2-3(6\left)\right)$
$\Rightarrow (\frac{p}{p+1}{)}^3+(\frac{q}{q+1}{)}^3+(\frac{r}{r+1}{)}^3=44$