Question

Show that: ${(x+y)}^7-x^7-y^7=7xy(x+y){(x^2+xy+y^2)}^2$.

Answer 1

The expression, $E$, on the left vanishes when $x=0$, $y=0$, $x+y=0$, hence it must contain $xy(x+y)$ as a factor.

Putting $x=\omega y$, we have

$E=\{{(1+\omega )}^7-{\omega }^7-1\}{y}^7=\{{(-{\omega }^2)}^7-{\omega }^7-1\}{y}^7$

$=(-{\omega }^2-\omega -1)y^7=0$;

Hence, $E$ contains $x-\omega y$ as a factor and similarly we may show that it contains $x-{\omega }^{2}y$ as a factor; that is, $E$ is divisible by

$(x-\omega y)(x-{\omega }^{2}y)$ or $x^2+xy+y^2$.

Further, $E$ being of degree seven, and $xy(x+y)(x^2+xy+y^2)$ of five dimensions, the remaining factor must be of the form $A(x^2+y^2)+Bxy$.

Thus,

${(x+y)}^7-x^7-y^7=xy(x+y)(x^2+xy+y^2)(A{x}^2+Bxy+A{y}^2)$.

Putting $x=1$, $y=1$, we have $21=2A+B$;

Putting $x=2$, $y=-1$, we have $21=5A-2B$; where $A=7$, $B=7$;

$\therefore {(x+y)}^7-x^7-y^7=7xy(x+y){(x^2+xy+y^2)}^2$.