How to prove that there exist α, β, γ ∈ R such that $αa^{15} + βb^{16} + γc^{17} = 1.$ Let R be a commutative ring with unity. Let $ a, b, c ∈ R $   be such
that there exist $x, y, z ∈ R$ with 
$xa + yb + zc = 1$
Show that there exist α, β, γ ∈ R such that
$$αa^{15} + βb^{16} + γc^{17} = 1.$$
MY trial  : I take $αa=βb= γc= \frac{1}{3}$ where $a= b= c=1$
as I don't know how to prove  it theoretically .
Pliz  help me  and tell me, 
thanks in advance 
 A: You know that $xa+by+zc=1$, now consider $$(ax+by+cz)^{46}=1$$ when expanding this product we get elements on the form $(ax)^i(by)^j(cz)^k$ where $i+j+k=46$. Since $i,j,k\geq 0$, all of the terms will have a factor of either $(ax)^{15}$, $(by)^{16}$, or $(cz)^{17}$. Thus you can write the expression as $$\alpha a^{15}+\beta b^{16}+\gamma c^{17}=1$$
Why $46$?
The worst case scenario is when $i=14$, and $j=15$, because then neither $a^{15}$ nor $b^{16}$ is factor of the term, and we want this product then to have a factor of $c^{17}$, that is $j=17$. So $i+j+k=14+15+17=46$.
A: Hint:
Expand $(xa+yb+zc)^{14+15+16+1}=1$ by the multinomial formula: it is the sum
$$\sum_{r+s+t=46}\frac{46!}{r!\,s!\,t!}\,x^ra^r\,y^sb^s\,z^tc^t. $$
and prove  that in each term, at least one of the inequalities
$$  r\ge 15,\quad s\ge 16,\quad t\ge 17, $$
is true, and group  terms for which you can factor out $x^{15}$, then  among the remaining terms, group those for which you can factor out $y^{16}$. The other terms will be factorable by $z^{17}$.
