A child's play

Without using the hint, simplifying the given expression, we get $\displaystyle x^{40} + 2x^{24} + 2x^{16} + x^8 + 4 + \frac{1}{x^8} + \frac{2}{x^{16}} + \frac{2}{x^{24}} + \frac{1}{x^{40}.}$ Now, let [ $n$ ] denote $x^n + \frac{1}{x^n}$ . We know that $\displaystyle [2^k] = [2^{k - 1}]^2 - [2]$ when $k$ is a positive integer and $k \ge 1$ . Thus, since $[1] = \sqrt{3}$ , then we have the following values: $\displaystyle [2] = 1, [4] = -1, [8] = -1, [16] = -1, [32] = -1.$ We are left with [ $40$ ] and [ $24$ ], but $[8] \times [16] = [24] + [8] = 1 \longrightarrow [24] = 2.$ Moreover, we get $[32] \times [8] = [40] + [24] = 1 \longrightarrow [40)] = -1.$ Indeed, plugging all the values above to the original expression, we finally have $-1 + 2(2) + 2( -1) -1 + 4 = \boxed{4}.$

$\frac { (x^{ 40 }+x^{ 24 }+x^{ 16 }+1)^{ 2 } }{ (x^{ 20 })^{ 2 } } =(\frac { x^{ 40 } }{ x^{ 20 } } +\frac { x^{ 24 } }{ x^{ 20 } } +\frac { x^{ 16 } }{ x^{ 20 } } +\frac { 1 }{ x^{ 20 } } )^{ 2 }=({ x }^{ 20 }+\frac { 1 }{ x^{ 20 } } +x^{ 4 }+\frac { 1 }{ x^{ 4 } } )^{ 2 }\\ x+\frac { 1 }{ x } =\sqrt { 3 } \rightarrow (x+\frac { 1 }{ x } )^{ 2 }=3\rightarrow { x }^{ 2 }+2+\frac { 1 }{ x^{ 2 } } =3\rightarrow x^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =1\\ (x^{ 2 }+\frac { 1 }{ x^{ 2 } } )^{ 2 }=1\rightarrow { x }^{ 4 }+2+\frac { 1 }{ x^{ 4 } } =1\rightarrow \boxed { { x }^{ 4 }+\frac { 1 }{ x^{ 4 } } =-1 } \\ (x^{ 4 }+\frac { 1 }{ x^{ 4 } } )^{ 2 }=1\rightarrow { x }^{ 8 }+2+\frac { 1 }{ { x }^{ 8 } } =1\rightarrow { x }^{ 8 }+\frac { 1 }{ x^{ 8 } } =-1\\ (x^{ 8 }+\frac { 1 }{ x^{ 8 } } )({ x }^{ 4 }+\frac { 1 }{ x^{ 4 } } )=1\rightarrow x^{ 12 }+x^{ 4 }+\frac { 1 }{ x^{ 4 } } +\frac { 1 }{ x^{ 12 } } =1\rightarrow x^{ 12 }+\frac { 1 }{ x^{ 12 } } =2\\ (x^{ 12 }+\frac { 1 }{ x^{ 12 } } )(x^{ 8 }+\frac { 1 }{ x^{ 8 } } )=-2\rightarrow x^{ 20 }+x^{ 4 }+\frac { 1 }{ x^{ 4 } } +\frac { 1 }{ x^{ 20 } } =-2\rightarrow \boxed { { x }^{ 20 }+\frac { 1 }{ x^{ 20 } } =-1 } \\ ({ x }^{ 20 }+\frac { 1 }{ x^{ 20 } } +x^{ 4 }+\frac { 1 }{ x^{ 4 } } )^{ 2 }=(-1-1)^{ 2 }=\boxed { 4 }$