Are the elements of $R=\mathbb{Q}[x]/\left<(x-1)^3\right>$ just constants?

Question

My understanding of the ring are that you treat the ideal as doing (x-1)^3=0 and therefore substituting 1 for x in every element, but it doesn’t make sense in the problem that I’m doing.

Answer 1

No, in this ring the polynomial $(x-1)^3$ is like $0$, in other words, $$ 0 = (x-1)^3 = x^3-3x^2+3x-1 \iff x^3 = 3x^2-3x+1 $$

Answer 2

My understanding of the ring are that you treat the ideal as doing $(x-1)^3=0$ and therefore substituting $1$ for $x$ in every element, but it doesn’t make sense in the problem that I’m doing.

Your understanding here is wrong. Here the equality $(x-1)^3=0$ is not an equality for a real number $x$, but an equality of polynomials, formal expressions in terms of the indeterminate $x$. So from $(x-1)^3=0$ you cannot conclude that $(x-1)=0$ and hence $x=1$; all you can conclude is that $x^3-3x^2+3x-1=0$, in other words that $x^3-3x^2+3x-1$ is considered "the same as" the $0$ polynomial in this ring.