Find solution of a Polynomial Equation

Question

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:

Answer 1

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 $A = -\frac{1}{4}$.

Answer 2

$$4b^2 + \frac{1}{b^2} -16 =0$$

$$4b^4 - 16b^2 + 1 = 0$$

$$let \;u = b^2$$

$$4u^2 - 16u + 1 = 0$$

$$u = \frac{4 \pm \sqrt{15}}{2}$$

$$b^2 = \frac{4 \pm \sqrt{15}}{2}$$

$$b = \sqrt{\frac{4 + \sqrt{15}}{2}}, \sqrt{\frac{4 - \sqrt{15}}{2}}, -\sqrt{\frac{4 + \sqrt{15}}{2}}, -\sqrt{\frac{4 - \sqrt{15}}{2}}$$

Then substitute into $b^4 + \frac{4}{b^4} - \frac{63}{b^2}$ as required.