Question

Let $p,q,r$ be the roots of $x^3+2x^2+3x+3=0$, then which of following is/are correct?

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

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{p}{p+1}+\frac{q}{q+1}+\frac{r}{r+1}=5$
B ${\left(\frac{p}{p+1}\right)}^3+{\left(\frac{q}{q+1}\right)}^3+{\left(\frac{r}{r+1}\right)}^3=44$

$x^3+2x^2+3x+3=0$ has roots as $p,q,r$
Let $\frac{p}{p+1}=y$, so
$\Rightarrow p=py+y\Rightarrow p=\frac{y}{1-y}$

As $p$ is a root of the given equation, so
$p^3+2p^2+3p+3=0\Rightarrow {\left(\frac{y}{1-y}\right)}^3+2{\left(\frac{y}{1-y}\right)}^2+3\left(\frac{y}{1-y}\right)+3=0\Rightarrow y^3+2y^2(1-y)+3y(1-y{)}^2+3(1-y{)}^3=0\Rightarrow y^3-5y^2+6y-3=0$
Whose roots are $y_1=\frac{p}{p+1},y_2=\frac{q}{q+1},y_3=\frac{r}{r+1}$

Now,
Sum of roots is
$\frac{p}{p+1}+\frac{q}{q+1}+\frac{r}{r+1}=5$

${\left(\frac{p}{p+1}\right)}^3+{\left(\frac{q}{q+1}\right)}^3+{\left(\frac{r}{r+1}\right)}^3=(y_1{)}^3+(y_2{)}^3+(y_3{)}^3=3y_1{y}_2{y}_3+(\sum y_1\left)\right[(\sum y_1{)}^2-3\sum y_1{y}_2]=3\cdot 3+\left(5\right)(5^2-3\cdot 6)=44$