# Third degree ploynomial question : 2

• Feb 28th 2009, 11:03 AM
champrock
Third degree ploynomial question : 2
if a,b,c are the roots of the equation: x^3 + 2x^2 + 3x + 3 = 0 then the value of

[a/(a+1)]^3 + [b/(b+1)]^3 + [c/(c+1)]^3 is?
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I know that this equation has one real root and two imaginary roots. the real root lies between -2 and 0. But unable to proceed further.
• Feb 28th 2009, 01:30 PM
red_dog
Let $\displaystyle y_1=\frac{a}{a+1}, \ y_2=\frac{b}{b+1}, \ y_3=\frac{c}{c+1}$

We form the equation with the roots $\displaystyle y_1, \ y_2, \ y_3$

$\displaystyle y=\frac{x}{x+1}\Rightarrow x=\frac{y}{1-y}$

Replace x in the x-equation an we get the equation in y: $\displaystyle y^3-5y^2+6y-3=0$

$\displaystyle S_1=y_1+y_2+y_3=5$

$\displaystyle S_2=y_1^2+y_2^2+y_3^2=(y_1+y_2+y_3)^2-2(y_1y_2+y_1y_3+y_2y_3)=25-12=13$

Let $\displaystyle S_3=y_1^3+y_2^3+y_3^3$

We have

$\displaystyle y_1^3-5y_1^2+6y_1-3=0$
$\displaystyle y_2^3-5y_2^2+6y_2-3=0$
$\displaystyle y_3^3-5y_3^2+6y_3-3=0$

Adding the three relations we have

$\displaystyle S_3-5S_2+6S_1-9=0\Rightarrow S_3=5S_2-6S_1+9=65-30+9=44$
• Feb 28th 2009, 07:39 PM
champrock
thanks for that solution.

can u please tell me some good resource on the net to practice solved question on third degree polynomials? I cant find good resources..