= Solution
{c}
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> $I=(X^3)_T=X^3T$ is precisely
$$
I=\left\{a_3X^3+\sum_{j\geq5}a_jX^j:\text{only finitely many }a_j\ne0\right\}.
$$
Indeed, 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 $I$, so
$$
\boxed{p+I=a+bX^2+cX^4+I}.
$$
For 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 $a=a'$, $b=b'$, $c=c'$.
In particular, $X^4\notin I$. Treating $I$ as the <principal ideal> of $\mathbb R[X]$ instead would incorrectly discard this surviving term.
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