<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] As $a$ is root of $f = X^3 - 12X + 8=0, a^3-12a+8=0$
Now, $b=\frac{a^2}2-4\implies a^2=2(b+4)=2b+8$
So, from  $a^3-12a+8=0, a(2b+8)-12a+8=0\implies a=-\frac8{2b+4}=-\frac4{b+2}$
So, $$\left(-\frac4{b+2}\right)^3-12\cdot \left(-\frac4{b+2}\right)+8=0$$
On simplification we  get , $$b^3-12b+8=0$$
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