Question

Prove that ${(a+b)}^5-a^5-b^5=5ab(a+b)(a^2+ab+b^2)$.

Answer 1

L.H.S$=(a+b{)}^5-a^5-b^5$

$=a^5+5a{b}^4+10a^2{b}^3+10a^3{b}^2+5a^{4}b+b^5-a^5-b^5$

$=5a{b}^4+10a^2{b}^3+10a^3{b}^2+5a^{4}b$

$=5ab(b^3+2a{b}^2+2a^{2}b+a^3)$

$=5ab\left\{\right(a+b\left)\right(a^2-ab+b^2)+2ab(a+b\left)\right\}$

$=5ab(a+b)(a^2-ab+b^2+2ab)$

$=5ab(a+b)(a^2+ab+b^2)$

$=$R.H.S

Alternate Method

Denote the expression on the left by $E$; then $E$ is a function of $a$ which vanishes when $a=0$; hence $a$ is a factor of $E$; similarly $b$ is a factor of $E$. Again $E$ vanishes when $a=-b$, that is $a+b$ is a factor of $E$; and therefore $E$ contains $ab(a+b)$ as a factor. The remaining factor must be of two dimensions, and, since it is symmetrical with respect to $a$ and $b$, it must be of the form $A{a}^2+Bab+A{b}^2$; thus
${(a+b)}^5-a^5-b^5=ab(a+b)(A{a}^2+Bab+A{b}^2)$,
where $A$ and $B$ are independent of $a$ and $b$.
Putting $a=1$, $b=1$, we have $15=2A+B$;
putting $a=2$, $b=-1$, we have $15=5A-2B$;
Hence $A=5$, $B=5$; and thus the required result at once follows.