<s>[INST] 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
 [/INST] 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
$$\begin{align}
{x^6+1\over x^3}&={(x^2+1)(x^4-x^2+1)\over x\cdot x^2}\\
&=\left(x+{1\over x} \right)\left(x^2-1+{1\over x^2} \right)\\
&=\left(x+{1\over x} \right)\left(\left(x+{1\over x}\right)^2-3\right)\\
&=\pm\sqrt{13}(13-3)=\pm10\sqrt{13}
\end{align}$$
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