Is it possible to solve this identity by "inspection"? 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.

$$(a+b)^3(a^5+b^5)+5ab(a+b)^2(a^4+b^4)+15a^2b^2(a+b)(a^3+b^3)+35a^3b^3(a^2+b^2)+70a^4b^4=(a+b)^8$$
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 $(a+b)^6$
