existence of $\alpha,\beta,\gamma\in R$ such that $\alpha a^{15} +\beta b^{16}+\gamma c^{17}=1$. Let $R$ be a commutative ring with identity. Let $a,b,c\in R$ such that $\exists $ $x,y,z\in R$ that satisfy $ax+by+cz=1$.
Prove that there exist $\alpha,\beta,\gamma\in R$ such that 
\begin{equation*}
\alpha a^{15}+\beta b^{16}+\gamma c^{17}=1.
\end{equation*}
I have not made any progress in this problem as I don't know how to start it.  Please help.
 A: Trick: raise $(ax+by+cz)$ to some power $n$, in order that in every monomial of the trinomial expansion: 


*

*either the degree wih respect to $a$ is $\geq 15$, 

*or the degree with respect to $b$ is $\geq 16$, 

*or the degree with respect to $c$ is $\geq 17$. 


For such a purpose, $n=15+16+17=48$ is enough. 
Collect pieces according to the three previously outlined situations, then notice that $(ax+by+cz)^n=1$.
A: Raise $ax+y+cz$ to the $15+16+17-2$ power, and use the multinomial formula:
\begin{align*}1=(ax+by+cz)^{46}&=\!\!\!\sum_{i+j+k=46} \frac{46!}{i!\,j!\,k!}(ax)^i(by)^j(cz)^k \\
&=\!\!\!\sum_{i+j+k=46} \frac{46!}{i!\,j!\,k!}a^ib^jc^k(x^iy^jz^k )
\end{align*}
Now gather together all terms containing $c^k$ with $k\ge 17$ and factor out $c^{17}$, then among the  remaining terms, gather all terms that contain  $b^j$ with $j\ge 16$ and factor out $b^{16}$. What remains are terms in $a^ib^jc^k$ with 
$$\begin{cases}k\le 16\\j\le 15\\i+j+k=46\end{cases},\quad\text{whence}\quad i\ge46-31=15,$$
so grouping them, you can factor out $a^{15}$.
