Algebraic equation I mean if we change interval of "x" to: 
$k>1$ and $x\in(\sqrt[3]{k},\sqrt[5]{k^3})$. 
Define the $B(x)$ as a polynomial as follows;
$B(x)=-2x^7-3x^6k+2k^2x^4-k^3x^3+2k^3x^2+3xk^4-2xk^3+k^4$
I want to check that whether $B(x)$ is positive?
if not, for which values of $k$ it is positive? 
Do we have to repeat the method separately for:
$$ y:=\sqrt[3]k $$
then plot B(x,y) and find "y" 
and also by defining 
$$ z:=\sqrt[5]{k^3} $$
and plot B(x,z)
and find "z"? 
And finally finding intersection of the solution in terms of "k?"
 A: Let $y:=\sqrt[3]k$ and solve the equation
$$x^7+3x^6-2x^5-\left(y^3+1\right)x^4+\left(2y^3-3\right)x^2+1=0.$$
We easily draw 
$$y=\sqrt[3]{\frac{x^7+3x^6-2x^5-x^4-3x^2+1}{x^4-2x^2}}$$ which is the zero set.
That curve is asymptotic to $x=y$ and $x=\sqrt2$, and there are two roots which satisfy $1<x<y$ for every value of $y$ above the minimum. The minimum is obtained by canceling the derivative, which amounts to $$3x^9+6x^8-12x^7-24x^6+12x^5+10x^4-4x^2+4=0.$$
We have one root $x\approx1.77368776,y\approx2.97774722$, then $k\approx26.4036205$.
There is also the isolated solution $x=1,y=1,k=1$.

A: Edit: As pointed out in the discussion wrong region for minimization was taken. Following plots display minima of $B(x)$ on $0<x<k^{1/3}$ which is not what the problems is concerned with.
It turns out that for small values of $k$ $B(x)$ is not always positive. For very large values of $k$ again minimum of $B(x)$ becomes negative (qualitatively for moderate values of $x$ term $-k x^4$ eventually dominates as $k$ gets large enough).
There is a range of values of $k$ for which $B$ is positive. I couldn't find any simple expression for the boundaries of this region but it's approximately $1.245 < k <26.4$
Some illustrative plots below
$k$">
$k$">
