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Factor $$\sum_{abc} a(b^3-c^3)$$

The solution given is as follows:

If we consider it as $f(a)$ then when $a=b$, $f(b)= 0$; thus, $b$ is a factor of $f(a)$. And since the equation is cyclic, other two factors are also cyclic. So the factors are: $$(a-b)(b-c)(c-a)$$

Now, my first doubt is why are we ignoring the other factors. For example, for $f(a)$, $a=b$ and $a=c$, $f(b)=f(c)=0$; thus, for $f(a)$ the factors should be $(a-b)(a-c)$.

Thus the overall factor of our polynomial then comes out to be: $$-(a-b)^2(b-c)^2(c-a)^2$$

Why is this not considered?

Besides this doubt, the book further says, since our polynomial is in a degree $4$ and the factors are of degree $3$, we need an extra first degree cyclic expression which will be another factor. Which is $m(a+b+c)$, with $m$ being a constant.

The conclusion is that the factors of the given polynomial are $$(a-b)(b-c)(c-a)m(a+b+c)$$

How did we guessed this extra term just by knowing that our degree was off?

Blue
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Ruchi
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    It's not considered because the overall polynomial is degree 4. Also, the polynomial has $(a-b)(b-c)(c-a)$ as a factor already, so only choice is $(a+b+c)$ since there is only degree 1 left. – Hypernova Apr 25 '21 at 08:07

2 Answers2

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A different point of view: looking for a determinantal common expression, a method that often works.

If one knows Vandermonde determinants like this one

$$\det \begin{pmatrix} 1 &1 &1 \\ a &b &c \\ a^2 &b^2 &c^2\end{pmatrix}=(a-b)(b-c)(c-a) \\$$

the similitude is evident ; you just have to modify the last line:

$$\det \begin{pmatrix} 1 &1 &1 \\ a &b &c \\ a^3 &b^3 &c^3\end{pmatrix} \\$$

And now, it remains to expand this determinant in two different ways:

  • with respect to the second line for the initial expression

  • by using $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, giving:

$$\det \begin{pmatrix} 1 &0 &0 \\ a &b-a &c-a \\ a^3 &b^3-a^3 &c^3-a^3\end{pmatrix} =\det \begin{pmatrix} 1 &0 &0 \\ a &(b-a) &(c-a) \\ a^3 &(b-a)(b^2+ba+a^2) &(c-a)(c^2+ca+a^2)\end{pmatrix} $$

$$=(b-a)(c-a)(c^2+ca+a^2-b^2-ba-a^2) =(b-a)(c-a)(c-b)(c+b+a)$$ as desired.

One can find this expansion as well here.

ultralegend5385
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Jean Marie
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(1) The thing is, the powers can be any number i.e. we can generalise that as $(a-b)^m(b-c)^m(c-a)^m$ is a factor of our expression. But our expression is of degree-four and the above expression is of degree $3m$. So, the only possible thing is $m=1$.

(2) We just knew from the offset of degree, that there is a linear factor missing, which is of course, of the form $k_1a+k_2b+k_3c$. But, the term should be cyclic. So, $k_1=k_2=k_3$, which the book considers $m$. So, the factor is $ma+mb+mc=m(a+b+c)$.

Now, that our factorisation is complete we have $$\sum_{\text{cyc}}a(b^3-c^3)=m(a+b+c)(a-b)(b-c)(c-a)$$ You can see that the LHS has all coefficients $1$, so we must have $m=1$. This implies $$\boxed{\sum_{\text{cyc}}a(b^3-c^3)=(a+b+c)(a-b)(b-c)(c-a)}$$

Hope this helps. Ask anything if not clear.

ultralegend5385
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