Factor $(a+b)(b+c)(c+a)+abc$. I know this is a cyclic polynomial, but I don't know how to solve problems like this. What should I do?
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
The polynomial $(a+b)(b+c)(c+a)+abc$ is not only cyclic, but also symmetric in $a,b,c$ and has total degree $3$, so if it factors, it should have a linear factor with total degree $1$. Also, it is "likely" that the linear factor is also cyclic and/or symmetric. So, try something like $a+b+c$ (the simplest symmetric polynomial in $a,b,c$ with total degree $1$) as one of the factors. Then determine what the other factor should be.
Spoiler: $(a+b)(b+c)(c+a)+abc = (a+b+c)(ab+bc+ca)$
answered Jul 17, 2014 at 4:47
$\begingroup$
$\endgroup$
$$(a+b+c)(ab+ac+bc)$$ using the elementary symmetric functions.
answered Jul 17, 2014 at 4:49
$\begingroup$
$\endgroup$
2
Take for example $x=a$ for variable. Expand: your expression is $(b+c) x^2+((b+c)^2+bc)x+bc(b+c)$. Now find the roots (compute $\Delta$, etc), and factorise.
-
-
$\begingroup$ Perhaps I do not understand your question: I have used $(x+b)(x+c)=x^2+(b+c)x+bc$, then multiply by $b+c$ and added $xbc$. (Supplementary hint: do not expand $(b+c)^2$ in $\Delta$.) $\endgroup$– KelennerJul 18, 2014 at 4:42