Ta có: $x+\dfrac{1}{x}=a$Ta có:
$A=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=a^3-3a$
$B=\left(x^3+\dfrac{1}{x^3}\right)^2-2=(a^3-3a)^2-2$
$x^2+\dfrac{1}{x^2}=\left(x+\dfrac{1}{x}\right)^2-2=a^2-2$
$x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=(a^2-2)^2-2=a^4-4a^2+2$
$C=\left(x^3+\dfrac{1}{x^3}\right)\left(x^4+\dfrac{1}{x^4}\right)-x-\dfrac{1}{x}=(a^3-3a)(a^4-4a^2-2)-a$