<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] 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.
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