When I solve $(a-b)^2$ I take $aa-ab-ab+bb$ I do not use formulas at all because I only forget them. To solve the above example all i do is to multiply one variable or constant at the time but when I ask anyone or anything how to solve $(a-b)^3$ all they tell me is to use this or that formula etc... And I hate formulas because I only forget them. I want to learn how to calculate math not how to use math formulas. Since no one so far can answer my question I am now turning my hope to you.

How do I calculate $(a-b)^3$ without formulas? I am supposed to do $aaa+a(-2)(-2)$ and so on or what?

I really appreciate all the help I can get, thanks!

  • $\begingroup$ $(a-b)^3=(a-b)\cdot (a-b)^2$ $\endgroup$ – Roman83 May 21 '16 at 12:21
  • 2
    $\begingroup$ Why do you call that ‘solving’ instead of ‘ expanding’? There's no unknown to find. $\endgroup$ – Bernard May 21 '16 at 12:23
  • $\begingroup$ You need to use the formula $x\cdot(y-z)=xy-xz$ twice (or, some may say, three times), paired with the formula $uv=vu$. $\endgroup$ – user228113 May 21 '16 at 22:46
  • $\begingroup$ There is no real formula to $(a \pm b)^n$. To be honest, it is just common sense in a simplified equation. But it is easier to remember as opposed to going through the steps, especially when $n \geqslant 7$ because then it gets confusing (well at least I myself find that as my breaking point if I was to multiply and go through each step and then simplify afterwards) $\endgroup$ – Mr Pie Sep 12 '17 at 3:46

Arrange $(a-b)^3$ as $(a-b)(a-b)(a-b)$. Now sum over all ways of picking either $a$ or $-b$ from each factor. So you get $$aaa+(-b)aa+a(-b)a+aa(-b)+(-b)(-b)a+(-b)a(-b)+a(-b)(-b)+(-b)(-b)(-b)$$

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