Finding value of an expression If $x^2-3x-1=0$ then find the value of $(x^6+1)/x^3$ I tried to solve the quadratic but it became too complicated any way of doing this without a calculator
 A: Since $x^2=3x+1$, we have $x^3=3x^2+x=10x+3$.
In the same way, $\frac{1}{x^2}=1-\frac{3}{x}$ gives $\frac{1}{x^3}=\frac{1}{x}-\frac{3}{x^2}=\frac{10}{x}-3$, hence:
$$\frac{x^6+1}{x^3} = x^3+\frac{1}{x^3} = 10\left(x+\frac{1}{x}\right) = \pm 10\sqrt{13},$$
because $x+\frac{1}{x}$ is $\pm\sqrt{\Delta}$, since the product of the roots of $x^2-3x-1$ is $-1$ by Vieta's theorem.
A: $$x-\frac1x=3.$$
Then
$$\left(x-\frac1x\right)^2+3=x^2+1+\frac1{x^2}=12.$$
Then
$$\left(x^2+1+\frac1{x^2}\right)\left(x-\frac1x\right)=x^3-\frac1{x^3}=36.$$
Finally,
$$\left(x^3+\frac1{x^3}\right)^2=\left(x^3-\frac1{x^3}\right)^2+4=1300.$$
A: A variant on the answers already given:
$$\begin{align}
x^2-3x-1=0&\implies x-{1\over x}=3\\
&\implies x^2-2+{1\over x^2}=9\\
&\implies x^2+2+{1\over x^2}=13\\
&\implies x+{1\over x}=\pm\sqrt{13}
\end{align}$$
and
$$\begin{align}
{x^6+1\over x^3}&={(x^2+1)(x^4-x^2+1)\over x\cdot x^2}\\
&=\left(x+{1\over x} \right)\left(x^2-1+{1\over x^2} \right)\\
&=\left(x+{1\over x} \right)\left(\left(x+{1\over x}\right)^2-3\right)\\
&=\pm\sqrt{13}(13-3)=\pm10\sqrt{13}
\end{align}$$
A: following from Jack's simplification, since the roots are $\alpha$ and $-\frac1{\alpha}$ we have
$$
\alpha + \frac1{\alpha} = \pm\sqrt{(\alpha - \frac1{\alpha})^2+4}=\pm\sqrt{13}
$$
A: As the product of the roots is $uv=-1$, they are opposite inverses of each other and the requested expression is the difference of their cubes, $u^3-v^3$.
Then,
$$u^3-v^3=(u-v)(u^2+uv+v^2)=\pm\sqrt{(u+v)^2-4uv}((u+v)^2-uv)=\pm\sqrt{13}\cdot10.$$
