Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$

Solution 1:

Let $c = -(a+b)$. Consider following cubic polynomial having $a, b, c$ as roots:

$$t^3-At^2 + Bt - C \stackrel{def}{=} (t-a)(t-b)(t-c) \quad\text{ where }\quad \begin{cases} A &= a + b + c = 0\\ B &= ab+bc+ca = -(a^2+b^2+ab)\\ C &= abc = -ab(a+b) \end{cases}$$ When $t$ is one of $a, b, c$, we have $t^3 = C - Bt$. This implies

$$t^7 = (C-Bt)^2 t = C^2 t - 2CB t^2 + B^2t^3 = (C^2 - B^3)t + B^2C -2CB t^2$$

Substitute $t$ by $a,b,c$ and sum over them. Together with

$$\begin{align}a + b + c &= A = 0\\ a^2 + b^2 +c^2 &= (a+b+c)^2 - 2(ab+bc+ca) = A^2 - 2B = -2B \end{align}$$

we obtain $$\begin{align} &a^7 + b^7 + c^7 = (C^2-B^3)A + 3B^2C - 2CB(A^2 - 2B) = 7B^2C\\ \implies & (a+b)^7 - a^7 - b^7 = -7B^2C = 7ab(a+b)(a^2+ab+b^2)^2 \end{align} $$

Solution 2:

expanding the left-hand side we get $$7\,{a}^{6}b+21\,{a}^{5}{b}^{2}+35\,{a}^{4}{b}^{3}+35\,{a}^{3}{b}^{4}+ 21\,{a}^{2}{b}^{5}+7\,a{b}^{6} $$ and the right-hand side: $$ 7\,{a}^{6}b+21\,{a}^{5}{b}^{2}+35\,{a}^{4}{b}^{3}+35\,{a}^{3}{b}^{4}+ 21\,{a}^{2}{b}^{5}+7\,a{b}^{6} $$

Solution 3:

As Mr. Chip suggested there’s no need to drag around $a$ and $b$ here. So I'm gonna proceed in this fashion: $$(1+x)^7-1^7-x^7=7x(1+x)(1+x+x^2)^2$$ Now we divide by $(1+x)$ or we can factor it out: Note that various divisions like that may call into question the rigor here. But it should be as easy as pie for you (I'm leaving it for you). Well, we go on $$(1+x)^6-(1-x+x^2-x^3+x^4-x^5+x^6)=7x(1+x+x^2)^2$$ $$7x+14x^2+21x^3+14x^4+7x^5=7x(1+x^2+x^4+2x+2x^2+2x^3)$$ $$7x(1+2x+3x^2+2x^3+x^4)=7x(1+2x+3x^2+2x^3+x^4)$$ So we got our result slightly faster this way. My point is you should be very proficient even when dealing with complicated expressions including the knoweledge of Multinomial theorem, which I used here on autopilot for getting the square of the trinomial. And the formula for $1\pm x^n$ and all the related subtleties are supposed to be easy when we talk about Olympiads. Seeing here $1^7+x^7$ should ring a bell that we can factor out $1+x\,.\,$ Hope my advice might be of some use.

Solution 4:

Let $I = (a+b)^7-a^7-b^7$.

$$a^7+b^7 = (a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)$$

and

$$(a+b)^6 = a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$$

Then \begin{eqnarray*} I &=& (a+b)(7a^5b+14a^4b^2+21a^3b^3+14a^2b^4+7ab^5)\\ &=&7ab(a+b)(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\\ & =& 7ab(a+b)(a^2+ab+b^2)^2 \end{eqnarray*}