0
$\begingroup$

I tried solving an exercise and was unsuccessful in doing so. I checked the solution and it had this piece $a^3 + b^3 = (a+b)^3-3ab(a+b)$.

Another exercise also had a similar formula: $a^2+b^2=(a+b)^2 - 2ab$

I can see a similar pattern but it's pointless to study the pattern if I don't understand it at all. Is this some kind of special product?

$\endgroup$
  • 1
    $\begingroup$ Have you tried expanding the RHS? $\endgroup$ – Wyllich Aug 30 '17 at 14:29
  • 1
    $\begingroup$ See cube of a binomial. $\endgroup$ – Mauro ALLEGRANZA Aug 30 '17 at 14:30
  • 1
    $\begingroup$ For the second case: $(a+b)^2= a^2+b^2+2ab$. $\endgroup$ – Mauro ALLEGRANZA Aug 30 '17 at 14:31
  • 1
    $\begingroup$ You can try with some very simple examples: $(3+2)^2=5^2=25 \ne 13=3^2+2^2$ $\endgroup$ – Mauro ALLEGRANZA Aug 30 '17 at 14:51
1
$\begingroup$

expanding the right-hand side we get $$a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2=...$$

$\endgroup$
  • $\begingroup$ Finally I see it, so if we expand $(a+b)^3$ we see an $a^3$ and a $b^3$ plus a whole lot extra. So you remove the extra. To end up again with $a^3 + b^3$ $\endgroup$ – user3051847 Aug 30 '17 at 14:39
2
$\begingroup$

$(a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+3ab(a+b)+b^3.$

$\endgroup$
0
$\begingroup$

A different way of thinking:

To take $3$ balls from $a$ and $b$, that are either all $3$ $a$ or all $3$ $b$, you can take any combination from $(a+b)$, but you have to subtract the options where both $a$ and $b$ occur. There are 3 such variations, one place out of three has a different symbol.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.