Respuesta :

dalendrk
[tex]Use:(a+b)^3=a^3+3a^2b+3ab^2+b^3\ (*)\\\\x^3+\left(\dfrac{1}{x}\right)^3\\=\underbrace{x^3+3\cdot x^2\cdot\dfrac{1}{x}+3\cdot x\cdot\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{x}\right)^3}_{(*)}-3\cdot x^2\cdot\dfrac{1}{x}-3\cdot x\cdot\left(\dfrac{1}{x}\right)^2\\\\=\left(x+\dfrac{1}{x}\right)^3-3x-3\cdot\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)[/tex]

therefore

[tex]x^3+\left(\dfrac{1}{x}\right)^3=110\iff \left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=110\\\\subtitute\ t=x+\dfrac{1}{x}\\\\t^3-3t=110\ \ \ \ |subtract\ 110\ from\ both\ sides\\\\t^3-3t-110=0\\\\t^3-25t+22t-110=0\\\\t(t^2-25)+22(t-5)=0\ \ \ |use\ a^2-b^2=(a-b)(a+b)\\\\t(t-5)(t+5)+22(t-5)=0\\\\(t-5)[t(t+5)+22]=0\iff t-5=0\ or\ \underbrace{t^2+5t+22=0}_{no\ solution}\\\\t=5\iff x+\dfrac{1}{x}=5\\\\Answer:\boxed{x+\dfrac{1}{x}=5}[/tex]

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