They call it Newton's Sum?

Algebra Level 4

Consider $a,b,c$ as complex and irrational numbers which satisfy:

$a+b+c=1$

$a^{2}+b^{2}+c^{2}=4$

$a^{3}+b^{3}+c^{3}=9$

If $a,b,c$ are roots of monic polynomial $x^{3}+px^{2}+qx+r$ .

Find $p+q+r$

Note:

Answer correctly up to three digits.

The answer is -3.666.

2 solutions

Chew-Seong Cheong
Sep 16, 2015

By Vieta's formula, we have: $\begin{cases} p = -(a + b + c) = -1 \\ q = ab + bc + ca \\ r = -abc \end{cases}$

Using Newton's sums method, we have:

$\begin{aligned} a^2+b^2+c^2 & = (a+b+c)^2 - 2(ab+bc+ca) \\ 4 & = 1 - 2(ab+bc+ca) \\ ab+bc+ca & = - \frac{3}{2} \\ \Rightarrow q & = -\frac{3}{2} \end{aligned}$

$\begin{aligned} a^3+b^3+c^3 & = (a+b+c)(a^2+b^2+c^2) - (ab+bc+ca)(a+b+c) +3abc \\ 9 & = (1)(4) - \left(-\frac{3}{2} \right)(1) + 3abc \\ abc & = \frac{7}{6} \\ \Rightarrow r & = - \frac{7}{6} \end{aligned}$

$\Rightarrow p + q + r = -1 -\dfrac{3}{2} - \dfrac{7}{6} = -\dfrac{11}{3} \approx \boxed{-3.667}$

Nice!

Slight improvement: You can use the following identity to simplify your work,

$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc).$

Pi Han Goh - 5 years, 9 months ago

correct thats way easier used the same way sir

Kaustubh Miglani - 5 years, 8 months ago

Akshat Sharda
Sep 15, 2015

Even I call it $Newton's$ $Sum$ $\ddot \smile$

I also think so. Newton sum has a role in it.

Anshuman Singh Bais - 5 years, 9 months ago

They call it Newton's Sum?

The answer is -3.666.

2 solutions

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