Divisibility of higher power polynomials

Question

I am stuck with this problem suggested to me by a friend . The problem reads, The power of $x^2+xy+y^2$ by which the polynomial $(x+y)^7-x^7-y^7$ is divisible are ? I tried it by writing $x^7+y^7$ expansion and simplifying the polynomials in factors but that doesn't help as the resulting expression is ugly . Someone kindly suggest a method to solve this problem.

Answer 1

Convert this problem to a one-variable problem by dividing a suitable power of $\ x$. Consider $$\ f(t)={(1+t)}^{7}-1-{t}^{7}$$ Note that $\ f(\omega)=0$. This implies that $\ {t}^{2}+t+1$ divides $\ f(t)$. Now, if the second power of $\ {t}^{2}+t+1$ is also a factor of $\ f(t)$, then $\ f'(\omega)=0$. Can you proceed?

I'll continue. $$\ f'(t)=7{(1+t)}^{6}-7{t}^{6}$$ Putting $\ t=\omega$, we have $$\ f'(\omega)=0$$ Hence the second power of $\ {t}^{2}+t+1$ divides $\ f(t)$. If the third power also divides $\ f(t)$ then $$\ f''(\omega)=0$$ $$\ f''(t)=42\left({(1+t)}^{5}-t^5\right)$$ Note that $\ f''(\omega) \neq 0$. Hence, the highest power of $\ {t}^{2}+t+1$ that divides $\ f(t)$ is $\ 2$. Hope it helps!

Answer 2

By the Binomial theorem we obtain: $$(x+y)^7-x^7-y^7=7xy(x^5+3x^4y+5x^3y^2+5x^2y^3+3xy^4+y^5)=$$ $$=7xy(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4+3xy(x^2-xy+y^2)+5x^2y^2)=$$ $$=7xy(x+y)(x^4+2x^3y+3x^2y^2+2xy^3+y^4)=7xy(x+y)(x^2+xy+y^2)^2.$$