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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$

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$