<s>[INST] Show that $8x^4 −16x^3 +16x^2 −8x+k = 0$ has at least one non-real root for all real $k$. Find the sum of the non-real roots 
Show that $8x^4 −16x^3 +16x^2 −8x+k = 0$ has at least one non-real root for all real $k$. Find the sum of the non-real roots. 

Since this polynomial looks so symmetric, I think factoring it might help. We have that $8(x^4-2x^3+2x^2-x) = 8x(x-1)(x^2-x+1)= -k$. Then I'm not sure how to work with the non-real root part, but I think that $(x^2-x+1)$ may have something to do with it.
 [/INST] The I-take-no-clever-shortcuts approach:
I consider the polynomial $p(x)=8x^4-16x^3+16x^2-8x$.
I try to rewrite $p(x)$ as something of the form  $d(ax^2+bx+c)^2$ because if it is so, I know how to find the roots. The $d$ coefficient is there only because I like my leading coefficient of $x^4$ to be rational, and $8$ is not a square, but $8/2=4=2^2$ it is. So let me assume $d=2$, which is like to say I'm considering the coefficients of $p(x)/2$
Expanding $(ax^2+bx+c)^2$ I find 
$$a^2x^4 + 2 a b x^3 +(b^2 + 2 a c) x^2+2 b c x+c^2$$
it is not difficult to see if there is a choice (actually there are two with opposite signs) of $a,b,c$ that makes the above polynomial equal to the source one, minus the constant terms (that we will send to the right anyway).
Indeed we can immediately find that $a=\pm 2$. Let me try with $a=2$. From that I derive (from the coefficient of $x^3$) that $b=-2$ and then $c=1$.
All in all we have obtained $p(x)=2(2x^2-2x+1)^2 -2$ and thus 
that $p(x)=-k$ can be rewritten as 
$$
2(2x^2-2x+1)^2 = 2-k
$$
Now, it is not difficult to work out an expression for the roots of this equation. You simply consider the two equations (one for each sign)
$$
2x^2-2x+1 = \pm \sqrt{1-k/2}
$$
and solve for $x$ using the classic formula. At the end you obtain four solutions, namely
$$
x_{1,2,3,4}=\frac{1}{2}\left(1\pm\sqrt{-1\pm\sqrt{4-2k}}\right)
$$
From here, it is easy to answer the questions of the problem.
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