# Ideal generated by $\langle a, b, c \rangle$ and $\langle a^{15}, b^{16}, c^{17} \rangle$?

Let $R$ be a commutative rings with unity such that $\langle a, b, c \rangle = 1$. Then show that $\langle a^{15}, b^{16}, c^{17} \rangle = 1$.

If I work in $\mathbb{Z}$, I can argue(due to prime factorisation) that gcd has not changed. But for general ring, I have no clue on how to begin with. One idea which I though of is to write $$xa + yb + zc = 1$$ and multiply it $n$-times so that I could get appropriate $a^n$ terms, but the problem with this approach is that there are cross terms which needed to be taken care.

Your intuition is good. Moreover, if you raise your equation to a high enough power, all the cross terms will end up in the ideal generated by $\{a^{15},b^{16},c^{17}\}$.
Specifically, by the binomial theorem, you can write $$1=(xa+yb+zc)^{n}=\sum_{i+j+k=n}\left(x^iy^jz^k{n\choose k}{i+j\choose i}\right)a^ib^jc^k.$$ Let's ignore the coefficient, and focus on the monomial in $a,\,b$ and $c$. Note that if $n\geq 15+16+17$, then in every summand here, at least one of the following holds: $i\geq 15$ or $j\geq 16$ or $k\geq 17$. Thus, you could factor at least one of $a^{15}$ or $b^{16}$ or $c^{17}$ from each summand and then collect these terms to get some linear combination of them summing to $1$. Thus, just raise your equation to the power of $15+16+17=48$, and you're done!