<s>[INST] How to find the value of $x^3(x^3-18)$? If $x(x-3)=-1$ then the value of $x^3(x^3-18)$ will be ?
$A)1\quad\quad\quad\quad B)-1\quad\quad\quad\quad C)0 \quad\quad\quad\quad D)|-1|$
I tried solving $x(x-3)=-1$:
$x^2-3x+1=0\quad\text{hence}\quad x_{1,2}=\cfrac{3\pm\sqrt{5}}{2}$. But it seems that doesn't help a lot to find the value of $x^6-18x^3$. as an alternative way, from the quadratic we have $x^3=3x^2-x$. if we plug in it $x^3(x^3-18)$ we obtain $(3x^2-x)(3x^2-x-18)$ but I can't continue from here.
Also the options are strange because $|-1|=1$ so it seems the answer should be  $B$ or $C$.
 [/INST] Since $x^2=3x-1$, $x^3=3x^2-x=8x-3$, so$$x^3(x^3-18)=(8x-3)(8x-21)=64x^2-192x+63=192x-64-192x+63=-1,$$so $B$) is correct. Alternatively, use$$x^6-18x^3+1=(x^2-3x+1)(x^4+3x^3+8x^2+3x+1).$$
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