Algebra and roots of polynomial

By letting $y=x+\frac{1}{x}$ ,

$x^2+\frac{1}{x^2}=y^2-2$

$x^3+\frac{1}{x^3}=y^3-3y$

$x^3+x^2+x+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}=28$

$\Rightarrow y+y^2-2+y^3-3y=28$

$\Rightarrow y^3+y^2-2y-30=0$

Applying the Rational Root Theorem, we know that $y=3$ is a root of the polynomial,

$(y-3)(y^2+4y+10)=0\Rightarrow$ only $y=3$ satisfies the condition that $x$ is a real number

$x+\frac{1}{x}=3\Rightarrow x^2-3x+1=0\Rightarrow x=\frac{3\pm\sqrt{5}}{2}$