Find solution of a Polynomial Equation If $4b^2+1/b^2=16$ then how do I find the solution of $b^4+4/b^4-63/b^2$?
From $4b^2+1/b^2=16$, I got $$(2b+1/b)^2 = 12 \tag{1}$$ and $$(2b-1/b)^2 = 20 \tag{2}.$$
By solving equation (1), $$2b+1/b = 2\sqrt{3}$$ and by solving equation (2), $$2b-1/b = 2\sqrt{5},$$
but I couldn't find a way to proceed further.
4 options are given below. From these 4, I have to choose one as answer.
a) -1/4
b) -2
c) 3
d) 1/4
[Edited]
Thank you all for the suggestions you made.  Now I have found the answer. The following is how I came across the solution.
Solution:

 A: $$4b^2 + \frac{1}{b^2} -16 =0$$
$$4b^4 - 16b^2 + 1 = 0$$
$$let \;u = b^2$$
$$4u^2 - 16u + 1 = 0$$
$$u = \frac{4 \pm \sqrt{15}}{2}$$
$$b^2 = \frac{4 \pm \sqrt{15}}{2}$$
$$b = \sqrt{\frac{4 + \sqrt{15}}{2}}, \sqrt{\frac{4 - \sqrt{15}}{2}}, -\sqrt{\frac{4 + \sqrt{15}}{2}}, -\sqrt{\frac{4 - \sqrt{15}}{2}}$$
Then substitute into $b^4 + \frac{4}{b^4} - \frac{63}{b^2}$ as required.
A: There's surely an elementary method that doesn't require solving for $b$, but here's a method that doesn't require much cleverness:
Denote $u := b^2$. Then, clearing denominators gives that the first equation is equivalent to $$f(u) := 4 u^2 - 16 u + 1,$$ and if we denote the quantity we're trying to evaluate by $A := b^4 - 63 b^{-2} + 4 b^{-4}$, then rewriting in terms of $u$, rearranging, and again clearing denominators gives that $$g(u) := u^4 - A u^2 - 63 u + 4 = 0 .$$
Since $u$ must be a root of both $f$ and $g$, the resultant $\operatorname{Res}_u(f, g)$ of $f$ and $g$ must vanish, and that resultant is quadratic in $A$.

 Computing gives $\operatorname{Res}_u(f, g) = 16 A^2 + 8 A + 1 = (4 A + 1)^2$, so $A = -\frac{1}{4}$.

