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I was able to find a demonstration, starting from the right part,
I wanted to start from the left part, in order to check if I understood well my knowdlege,

Starting from left or right does make any difference in how it's complex the resolution? And, in case yes, what criteria should I use to identify which part is simpler?

My current steps are these:

(intuition, with the expanded form, I can have some chance to re-group as I need, I will add the missing term but with neutral values)

ab+bc+ca = ab(c+'c) + bc (a+'a) + ca (b+'b)
................ = abc + ab'c + abc + 'abc + abc + a'bc
................ = how should I proceed?

(the single quote is not)

The steps from the right are these:
(a+b)(b+c)(c+a) = a(b+c)(c+a)+b(b+c)(c+a)
........................... = abc+aab+acc+aac+bbc+abb+bcc+abc
........................... = abc+ab+ac+ac+bc+ab+bc
........................... = abc+ab+ac+bc
........................... = ab(c+1) + ac + bc
........................... = ab+ac+bc

Following the suggestion in the comment, i.e. how's supposed to deduct that starting from ab+ac+bc, I should add ab(c+1)? The general intuition here is what's missing, thank you

Delayer
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    Since this is an equality, just reverse the steps, read your proof from bottom to top. – MasB Feb 18 '24 at 15:24
  • @MasB True, but the reverse steps, seems very unnatural from the bottom (i.e. guessing I shoud add c+1? why?) , like suggesting there should be another way to solve it from the bottom. I miss the intuition in why I should have started right, basically. – Delayer Feb 18 '24 at 15:35
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    Distribution works both ways, also over AND: $$ab+bc+ca = (a+bc+ca)(b+bc+ca)$$ – peterwhy Feb 18 '24 at 16:00
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    Please use MathJax. Here is a tutorial. – Dietrich Burde Feb 18 '24 at 17:21

1 Answers1

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Distribution works both ways, also over AND. To mimic your attempt from right to left:

$$\begin{align*} ab+bc+ca &= (a+bc+ca)(b+bc+ca)\\ &= (a+b+c)(a+b+a)(a+c+c)(a+c+a)(b+b+c)(b+b+a)(b+c+c)(b+c+a)\\ &= (a+b+c)(a+b)(a+c)(a+c)(b+c)(a+b)(b+c)\\ &= (a+b+c)(a+b)(a+c)(b+c)\\ &= (a+b+c0)(a+c)(b+c)\\ &= (a+b)(a+c)(b+c) \end{align*}$$

peterwhy
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