$f(x)=x^3-21x+35$ is irreducable over $Q$. Show if $f(\alpha)=0$ then $f(\alpha^2+2\alpha-14)=0$ I am not sure how to do this in a short, nice way.
I have proved it by factoring $f(x)=(x-\alpha)$ and plugging in, but that is very very messy. Is there a slick way?
Thank you!

We can prove that is $\alpha$ is a root of $x^3-21x+35=0$ then $\alpha^2+2\alpha-14$ is a root of the same equation.
First use completion of squares to render
$\alpha^2+2\alpha-14=(\alpha+1)^2-15$
Now suppose that $\alpha^3-21\alpha+35=0$.  Define $A=\alpha+1$:
$(A-1)^3-21(A-1)+35=0$
$A^2-3A^2+3A-1-21A+21+35=0$
$A^3-3A^2-18A+55=0$
Next separate even and odd degree terms and square both sides:
$A^3-18A=3A^2-55$
$A^6-36A^4+324A^2=9A^4-330A^2+3025$
$B^3-45B^2+654B-3025=0$
where $B=A^2$.
Now define $B=C+15$ where $C=\alpha^2+2\alpha-14$ is your target function:
$(C+15)^3-45(C+15)^2+654(C+15)-3025=0$
$C^3+45C^2+675C+3375-45C^2-1350C-10125+654C+9810-3025=0$
And somewhat magically, when you combine like terms the big numbers mostly cancel:
$\color{blue}{C^3-21C+35=0}$
