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I was asked to prove the following identity (starting from the left-hand side): $$(a+b)³(a⁵+b⁵)+5ab(a+b)²(a⁴+b⁴)+15a²b²(a+b)(a³+b³)+35a³b³(a²+b²)+70a⁴b⁴=(a+b)^8.$$ I'm trying to solve it by a sort of "inspection", but I haven't made it yet. Of course I could try to expand the left-hand polynomial and come to a more recognizable form of $(a+b)^8$, but of course that would be the hard way (assuming that there is an easy one).

As an example of why I am talking of "inspection" I can state a similar problem:

Show that $$(x+\frac{5}{2}a)⁴-10a(x+\frac{5}{2}a)³+35a²(x+\frac{5}{2}a)²-50a³(x+\frac{5}{2}a)+24a⁴=(x²-\frac{1}{4}a²)(x²-\frac{9}{4}a²).$$ Here by "inspection" we can deduce that the left-hand side of the identity is equivalent to $$[(x+\frac{5}{2}a)-a][(x+\frac{5}{2}a)-2a][(x+\frac{5}{2}a)-3a][(x+\frac{5}{2}a)-4a]$$ and then after a few steps come to the the desire result.

I would appreciate any help you could give me.

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  • $\begingroup$ Perhaps you could use the sum of powers identity lots of times. $\endgroup$
    – cgss
    Aug 29, 2020 at 22:10

2 Answers 2

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$$(a+b)^3(a^5+b^5)+5ab(a+b)^2(a^4+b^4)+15a^2b^2(a+b)(a^3+b^3)+35a^3b^3(a^2+b^2)+70a^4b^4=(a+b)^8$$

Note that $a+b|(a+b)^3$, so except the last two terms in the LHS, every term is divisible by $(a+b)^2$, in fact you can take $35a^3b^3$ common from the last two terms to have $$35a³b³(a²+b²)+70a⁴b⁴=35a^3b^3(a+b)^2$$ so that the LHS becomes

$$(a+b)^2\times((a+b)(a^5+b^5)+5ab(a^4+b^4)+15a^2b^2(a^2-ab+b^2)+35a^3b^3)$$

and if you leave out $(a+b)^2$ from this now, the multiplying and expanding becomes easier, you can verify that it is indeed $(a+b)^6$

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  • $\begingroup$ Wow, this is an excellent approach. Thank you! $\endgroup$ Aug 29, 2020 at 22:29
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Since the expression is symmetric and in decreasing order, it suffices to consider only the expansion for $a$ that is

$$a^8+3a^7+3a^6+a^5+5a^7+10a^6+5a^5+15a^6+15a^5+35a^5+70a^4$$

$$a^8+8a^7+28a^6+56a^5+70a^4$$

which agrees with the row of Pascal's triangle for $n=8$ that is

$$(a+b)^8=\sum_{k=0}^8 \binom 8 k a^kb^{8-k}$$

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    $\begingroup$ How is this solving by "inspection"? $\endgroup$
    – amWhy
    Aug 29, 2020 at 22:07
  • $\begingroup$ @amWhy I've adjusted the terms and added an example. $\endgroup$
    – user
    Aug 29, 2020 at 22:09
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    $\begingroup$ Yes, but in order to do so I would have to expand the left-hand side of the identity, which is what I want to avoid; if possible. $\endgroup$ Aug 29, 2020 at 22:11
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    $\begingroup$ Nice trick, @user! $\endgroup$
    – cosmo5
    Aug 30, 2020 at 9:34
  • $\begingroup$ @cosmo5 Thanks! Indeed it seems not at all a bad method. Bye $\endgroup$
    – user
    Aug 30, 2020 at 9:35

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