 Solution

All principal ideals in this part are taken inside the subring $T$, rather than inside $\mathbb{R}[X]$. First, $T$ is a subring: sums and additive inverses retain zero coefficient of $X$, and the coefficient of $X$ in a product $uv$ is $u(0)[X]v+v(0)[X]u=0$. It contains $1$.

The principal ideal $I=(X^3{)}_T=X^{3}T$ is preciselyIndeed, multiplying an element of $T$ by $X^3$ gives a possible term of polynomial degree $3$ and terms of polynomial degrees at least $5$, with no $X^4$ term. Conversely, dividing any displayed polynomial by $X^3$ gives a polynomial with no $X$ term, hence an element of $T$.

Given $p\in T$, let $a,b,c$ be its coefficients of $1,X^2,X^4$, respectively. Subtracting these terms leaves an element of $I$, soFor uniqueness, if two such representatives differ by an element of $I$, their difference has possible monomials only $1,X^2,X^4$, whereas every element of $I$ has zero coefficients at all three. Comparing coefficients gives $a=a^{\prime }$, $b=b^{\prime }$, $c=c^{\prime }$.

In particular, $X^4\notin I$. Treating $I$ as the principal ideal of $\mathbb{R}[X]$ instead would incorrectly discard this surviving term.