Show that $c^2 + a^2d=abc$ for a monic quartic polynomial I apologize in advance for asking a homework question, but I have genuinely no idea on how to approach part b). The question is as follows:

Consider the polynomial equation $\rm{P}(x)=x^4 + ax^3 + bx^2 + cx+d$, with a root $ki$. In addition, $a,b,c,d\in Z$ and $k\in R,\; k\neq 0$
(a) Show that $c=k^2a$
(b) Show that $c^2 + a^2d=abc$

Part (a) is quite trivial and can be done by knowing that $ki$ and $\bar{ki}$ are roots since all the coefficients are real. However I am unsure of how to do part (b), do I square the expression in part (a). I tried working backwards from the solution to a true statement here is what I had:
$$
\begin{align}
c^2 + a^2d &=abc \\
k^4a^2+a^2d &=abc \\
a^2\left(k^4 +d\right) &=abc \\
a\left(k^4 +d\right) &=bc \tag{assuming $a\neq 0$, though this may not be true} \\
\dots &\dots \\
k^4 &=\frac{bc-ad}{a}
\end{align}
$$
This gets me nowhere

We know that $ki$ is a solution to $P(x)=0$ so in particular,
$$\begin{split}
(ki)^4+a(ki)^3+b(ki)^2+c(ki)+d&=0 \\
\Leftrightarrow k^4-ak^3i-bk^2+cki+d&=0 \\
\Leftrightarrow k^4-bk^2+d+ki(c-ak^2)&=0
\end{split}$$
Using part (a), this gives $k^4-bk^2+d=0$. Now add $bk^2$ on both sides and multiply through with $a^2$ to get $$a^2k^4+a^2d=a^2bk^2=abak^2.$$ Due to part (a) we can substitute $c^2$ for $a^2k^4$ as well as $c$ for $ak^2$ and obtain
 $$c^2+a^2d=abc.$$
