<s>[INST] How to resolve this algebra equation? $$f = X^3 - 12X + 8$$ $a $-  complex number, $a$ is a root for $f$
$b = a^2/2 - 4 $. 
Show that $f(b) = 0$
This is one of my theme exercises ... Some explanations will be appreciated ! Thank you all for your time .
 [/INST] \begin{align*}
f(b) &= f(\frac{a^2}{2} - 4) \\
&= (\frac{a^2}{2} - 4)^3 - 12(\frac{a^2}{2} - 4) + 8  \\
&= \frac{a^6}{8} - 3a^4 + 24 a^2 - 64 - 6a^2 + 48 + 8 \\
&= \frac{a^6}{8} -3a^4 + 18a^2 -8
\end{align*}
On the other hand,
$$
f(a) = a^3 - 12a + 8 = 0
$$
and therefore 
\begin{align*}
0 = 0^2 &= (a^3 - 12a + 8)^2 \\
&= a^6 - 24a^4 + 16a^3 + 144 a^2 - 192a + 64 \\
&= (a^6 - 24a^4 + 144 a^2 - 64) + (16a^3 - 192a + 128) \\
&= 8(\frac{a^6}{8} -3a^4 + 18a^2 -8) + 16(a^3 - 12a + 8)\\
&= 8f(b) + 16f(a) \\
&= 8f(b)
\end{align*}
Hence,
$$
f(b)=0.
$$
</s>