= Solution
{c}
Set $I=(X^3)_T$ and $J=(Y^3)_{\mathbb R[Y]}$. Part (b)(i) gives a unique representative $a+bX^2+cX^4$ for each element of the <quotient ring> $T/I$. By <polynomial division>, each element of the <quotient ring> $\mathbb R[Y]/J$ likewise has a unique representative $a+bY+cY^2$. Thus the proposed map $\Phi$ is well-defined and a <bijection>.
It preserves <addition> and $1$. To verify <multiplication>, all terms of <polynomial degree> at least $5$ lie in $I$. Hence
$$
\begin{aligned}
&(a+bX^2+cX^4)(a'+b'X^2+c'X^4)+I\\
&\qquad=aa'+(ab'+ba')X^2+(ac'+bb'+ca')X^4+I.
\end{aligned}
$$
In the other <quotient ring>, $Y^3=0$, giving exactly the same <coefficient> formula:
$$
(a+bY+cY^2)(a'+b'Y+c'Y^2)+J
=aa'+(ab'+ba')Y+(ac'+bb'+ca')Y^2+J.
$$
Therefore
$$
\boxed{\Phi:T/(X^3)_T\xrightarrow{\ \cong\ }\mathbb R[Y]/(Y^3),\qquad
a+bX^2+cX^4+I\longmapsto a+bY+cY^2+J}
$$
is a <ring isomorphism>. Both sides are examples of a <truncated polynomial ring>; the generator on the left is the class of $X^2$, whose square is nonzero and whose cube vanishes.
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