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I am at a loss at how to approach this problem. It doesn't make sense to me to solve without finding the roots. I couldn't find the roots without a calculator anyway, but you're not meant to.

Show that if the roots of the equation:

$5x^3-x^2-2x+3=0$

are a,b and c then:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2/3$

$a^2+b^2+c^2=\frac{21}{25}$

$a^3+b^3+c^3=-\frac{194}{125}$

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You'll need the Newton-Girard formulae along with the Vieta formulae for this problem. –  J. M. Aug 6 '12 at 13:45
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Why do you call this a "fundamental theorem of algebra type problem"? It really has nothing to do with algebraic closure of the complex numbers. –  Marc van Leeuwen Aug 6 '12 at 14:01
    
because that was one of the hints in the question. I have no idea why to be honest. I thought it was to do with the number of factors being equal to the number of the highest power –  Magpie Aug 6 '12 at 14:02
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3 Answers 3

up vote 2 down vote accepted

If you haven't yet been exposed to the general theory, there's nothing stopping you working from first principles:

Given

$$ 5x^3−x^2−2x+3=0 $$

we have

$$ x^3−(1/5)x^2−(2/5)x+3/5=0 $$

Now if $a$, $b$, $c$ are the roots, then

$$ (x-a)(x-b)(x-c) = 0 $$

which is to say

$$ x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0 $$

so we know

$$ \begin{align*} a+b+c &= 1/5\\ ab+bc+ca &= -2/5\\ abc &= -3/5 \end{align*} $$

which are enough to establish the things you're asked to establish, with some manipulation. For example,

$$ \begin{align*} 1/a + 1/b + 1/c &= bc/abc + ca / abc + ab / abc\\ &= (ab+bc+bc) / abc \end{align*} $$

and

$$ a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) $$

The sum of cubes is left as an exercise.

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Hey thanks! This looks a bit more friendly. I will have a go at it myself. –  Magpie Aug 6 '12 at 15:10
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What Aakash has shown is in fact the traditional way to derive the Vieta formulae. –  J. M. Aug 6 '12 at 15:19
    
ok thanks for letting me know. –  Magpie Aug 6 '12 at 17:06
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This is by no means "precalculus", and no calculator will help you to see that. Also, your $y = 0$ is nonsensical, except if you want to state the problem as "$a, b, c$ are roots to $y(x) = 0$, where $y(t) = 5t^3 ...$"

A simple application of Viete's relations and Newton's formulae solves most of such problems. For instance,

$$\sum \frac{1}{a} = \frac{ab + bc + ca}{abc} = \frac{-2/5}{-3/5} = \frac{2}{3}$$

$$a^2 + b^2 + c^2 = (a + b + c)^2 - 2\sum ab = \frac{1}{25} - \frac{2(-2)}{5} = \frac{21}{25}.$$

For the third one, try summing like $y(a) + y(b) + y(c) = 0$ and conclude something about the LHS.

Edit: Of course, you're talking about the Fundamental theorem of Symmetric Polynomials, stating that expressions such as yours (i.e., symmetric) can always be written in terms of the elementary symmetric polynomials appearing in Viete's relations.

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yeah you were right, that y, wasn't actually in the question. I changed it. –  Magpie Aug 6 '12 at 13:46
    
Of course? I am even more lost. viete is not something I have heard of. –  Magpie Aug 6 '12 at 14:03
    
Per chance the hint is about the fundamental theorem of symmetric polynomials, instead of being about algebra? –  awllower Aug 6 '12 at 14:34
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@AakashM Indeed, I had lost myself a bit there. From another point of view, the background is that for $S = \mathbb{C}[x_1, \ldots, x_n]$, $S_n$ acts on $S$ by permuting indeterminates and fixes the subalgebra $\mathbb{C}[\sigma_1, \ldots, \sigma_n] \simeq S$. –  Chindea Filip Aug 6 '12 at 14:59
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The hint referring to the Fundamental Theorem of Algebra was to get you to realize that a cubic polynomial will have three complex roots. In other words, we can write your polynomial as $$ y=x^3-\frac{1}{5}x^2-\frac{2}{5}x+\frac{3}{5} $$ and as $$ y=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc $$ so equating terms with like exponents we have

$a+b+c=\frac{1}{5}$

$ab+ac+bc=-\frac{2}{5}$

$abc=-\frac{3}{5}$

Now we can solve your problems.


(1) Using the values above we have $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{ab+ac+bc}{abc} = \frac{-2/5}{-3/5}=\frac{2}{3} $$


(2) From $$ (a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc) $$ we have $$ \left(\frac{1}{5}\right)^2 = (a^2+b^2+c^2)+2\left(-\frac{2}{5}\right) $$ so $a^2+b^2+c^2=21/125.$


(3) It'll be easiest to use a preliminary result for this part. From $$ (a+b+c)(ab+ac+bc) = (a^2b+ab^2+a^2c+ac^2+b^2c+bc^2) + 3abc $$ it's not hard to show that $$ a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=\frac{43}{25} $$ So from $$ (a+b+c)^3 = (a^3+b^3+c^3)+(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+2(a+b+c)(ab+ac+bc) $$ we can conclude $$ a^3+b^3+c^3 = -\frac{194}{125} $$

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Your answer in combination with the chosen one, is very helpful thanks. –  Magpie Aug 6 '12 at 16:59
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