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I just stumbled upon a question to figure out how to simplify

J = (ab)+(ac)-(b+c)

My steps:

<=> a*(b+c)-b-c

<=> a*(b+c) -1*(b+c)

But that was not one of the solutions. One of these was, as mentioned above, (a-1)*(b+c).

As I saw this I somehow knew it is correct, calculated it and yes it is. But my math is a bit outdated and I cannot remember the law to see this. I do know it is correct, but by the love of god, I still don't know how to pull it of.

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  • $\begingroup$ You are almost done. $a(b+c) -1(b+c)$ can be factored again (b+c) and will yield: $(a-1)(b+c)$ $\endgroup$ – Dashi Nov 26 '18 at 18:42
  • $\begingroup$ Your title has $+$ instead of $*$ $\endgroup$ – Ross Millikan Nov 26 '18 at 18:54
  • $\begingroup$ aaah, now I see it. It is not any fancy rule, it is just another factorization. Something tree, something forest.. Thank you!! $\endgroup$ – InDaPond Nov 26 '18 at 18:57
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It is called the distributive law $$ (x+y)z=xz+yz. $$ You apply it backwards in $$ (a-1)\underbrace{(b+c)}_{z}=a\underbrace{(b+c)}_{z}-1\underbrace{(b+c)}_{z}. $$

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The first two terms are $a(b+c)$ by distributivity of multiplication over addition. A reuse of this rule, together with $x=1x$ with $x=b+c$, completes the proof.

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