(x+1/x)^2 = 3 expand the left side
x^2 + 2 + 1/x^2 = 3 subtract 3 from both sides
x^2 - 1 + 1/x^2 = 0 (1)
Now x^3 + 1/x^3 is a sum of cubes which can be factored as
(x + 1/x) ( x^2 - 1 + 1/x^2 ) sub (1) into this factorization
( x + 1/x) ( 0 ) =
0
If (x+1/x)^2=3, what is x^3+1/x^3
\(\begin{array}{|rcll|} \hline \left(x+\frac{1}{x}\right)^2 &=&3 \\ x+\frac{1}{x} &=& \sqrt{3} \\ \hline \end{array} \)
\(\small{ \begin{array}{|rclcll|} \hline (x+\frac{1}{x})^3 &=& \left(x+\frac{1}{x}\right)^2 \cdot \left(x+\frac{1}{x}\right) &=& 3\cdot \sqrt{3} \\ x^3+3x^2\cdot \frac{1}{x} +3x\cdot \frac{1}{x^2}+\frac{1}{x^3} &=& && 3\cdot \sqrt{3} \\ x^3+3x +3\cdot \frac{1}{x}+\frac{1}{x^3} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3}+ 3x +3\cdot \frac{1}{x} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3}+ 3\cdot(x + \frac{1}{x}) &=& && 3\cdot \sqrt{3} \quad & | \quad x + \frac{1}{x}=\sqrt{3} \\ x^3+\frac{1}{x^3}+ 3\cdot\sqrt{3} &=& && 3\cdot \sqrt{3} \\ x^3+\frac{1}{x^3} &=& && 3\cdot \sqrt{3}- 3\cdot\sqrt{3} \\ \mathbf{x^3+\frac{1}{x^3}} &\mathbf{=} & &&\mathbf{ 0 } \\ \hline \end{array} }\)