Question

If $x^3+\frac{1}{x^3}=2$ then the value of $\frac{x^2+1}{x}$ is

• A. $-1,2$
• B. $1,2$
• C. $-1,-2$
• D. $1,-2$

Answer 1

The correct option is A $-1,2$

$(x+\frac{1}{x}{)}^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$

Let $x+\frac{1}{x}=y$

Then $y^3=2+3y$

$⟹y^3-3y-2=0$

For $y=1,1-3-2\ne 0y=2,8-6-2=0$

$⟹2$ is a factor

$(y-2)(y^2+2y+1)=0$

$⟹(y-2)(y-1)=0$

$⟹y=-1,2$

$x+\frac{1}{x}=-1,2$

$⟹\frac{x^2+1}{x}=-1,2$