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I want to factor $$bc^2+ab^2+a^2c-b^2c-ac^2-a^2b$$

Using Wolfram, I know it's factored into $$-(a-b)(a-c)(b-c) = (b-a)(a-c)(b-c)$$

However, I don't think I ever got taught how to simplify such expression by hand.

  • What's the general methodology for this?

  • Is there any mental-tool that could be used to do this without the experience to spot factors by sight?

B. Lee
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    The power on $a$ appears to be in error. I believe that you meant $a^2c$. I tried to edit it, but it's too small an edit for me to perform. – Michael Burr Mar 01 '15 at 19:55

1 Answers1

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This is an expression deduced from a pattern by a circular permutation on the variables; it is more apparent if you rewrite it as: $$ab^2+bc^2+ca^2- a^2b-b^2c-c^2a$$ To (try to) factor it, a strategy consists in breaking the symmetry between variables, e.g. by replacing $b-a$ with $(b-c)+(c-a)$:

\begin{align*} ab^2+bc^2+ca^2- a^2b-b^2c-c^2a&= ab(b-a)+bc(c-b)+ca(a-c)\\ &=ab(b-c)+ab(c-a)+bc(c-b)+ca(a-c)\\ &=(ab-bc)(b-c)+(ab-ca)(c-a)\\ &=b(a-c)(b-c)+a(b-c)(c-a)\\ &=(a-b)(b-c)(c-a). \end{align*}

Bernard
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