Cho $f\left(x\right)=\left(x^3+12x-31\right)^{2013}$. Tính $f\left(a\right)=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}$.

Đặt \( a = \sqrt[3]{16 - 8\sqrt{5}} + \sqrt[3]{16 + 8\sqrt{5}} \). Gọi \( u = \sqrt[3]{16 - 8\sqrt{5}} \) và \( v = \sqrt[3]{16 + 8\sqrt{5}} \). Ta có: \[ a = u + v \] Cubing both sides, we get: \[ a^3 = (u + v)^3 \] \[ a^3 = u^3 + v^3 + 3uv(u + v) \] Since \( u^3 = 16 - 8\sqrt{5} \) and \( v^3 = 16 + 8\sqrt{5} \), we have: \[ u^3 + v^3 = (16 - 8\sqrt{5}) + (16 + 8\sqrt{5}) = 32 \] Next, we need to find \( uv \): \[ uv = \sqrt[3]{(16 - 8\sqrt{5})(16 + 8\sqrt{5})} \] \[ uv = \sqrt[3]{16^2 - (8\sqrt{5})^2} \] \[ uv = \sqrt[3]{256 - 320} \] \[ uv = \sqrt[3]{-64} \] \[ uv = -4 \] Now substitute back into the equation for \( a^3 \): \[ a^3 = 32 + 3(-4)(a) \] \[ a^3 = 32 - 12a \] \[ a^3 + 12a - 32 = 0 \] We need to solve this cubic equation. By inspection or using rational root theorem, we find that \( a = 2 \) is a solution: \[ 2^3 + 12(2) - 32 = 0 \] \[ 8 + 24 - 32 = 0 \] \[ 0 = 0 \] Thus, \( a = 2 \). Finally, we need to compute \( f(a) \): \[ f(a) = (a^3 + 12a - 31)^{2013} \] \[ f(2) = (2^3 + 12(2) - 31)^{2013} \] \[ f(2) = (8 + 24 - 31)^{2013} \] \[ f(2) = (1)^{2013} \] \[ f(2) = 1 \] Therefore, the value of \( f(a) \) is: \[ \boxed{1} \]