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What is the absolute difference between the two real numbers $x$ for which $(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$? Express your answer in simplest radical form

I tried guessing solutions but seeing how there are no common zeroes to both the left- and right-hand sides I don't know what to do.

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    $\begingroup$ One (perhaps inelegant) approach is to expand both sides by distributing; the terms $x^3$ cancel, and you're left with a quadratic equation in $x$. $\endgroup$ – user296602 Dec 11 '15 at 4:53
  • $\begingroup$ That may work. This question was given as a question to students to solve in under $45$ seconds. $\endgroup$ – Puzzled417 Dec 11 '15 at 4:53
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    $\begingroup$ It'll be far faster if you take care of those differences of squares on both sides $\endgroup$ – David Peterson Dec 11 '15 at 4:54
  • $\begingroup$ Can you show me what you mean? $\endgroup$ – Puzzled417 Dec 11 '15 at 4:55
  • $\begingroup$ I don't know if it helps to solve it in less than $45$ seconds, but you may remark that the absolute difference of the real roots of a quadratic function $ax^2+bx+c$ is $\frac{\sqrt{\Delta}}{a}$, where $\Delta=b^2-4ac$ is the discriminant. $\endgroup$ – Taladris Dec 11 '15 at 5:03
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We have $$(x^2-1)(x-2) = (x^2-9)(x+2) \implies x^3-2x^2-x+2 = x^3+2x^2-9x-18$$ which simplifies to $$4x^2-8x-20 = 0 \implies x^2-2x-5 = 0 \implies (x-1)^2 = 6$$ Hence, the roots are $1\pm\sqrt6$, which means that the difference between the roots is $2\sqrt6$.

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$(x^2-1)(x-2)-[(x^2-9)(x+2)]=0$ thus $x^3-x-2x^2+2-x^3+9x-2x^2+18=0$ so $-4x^2+8x+20=0$ thus Solving we get $x=1\pm \sqrt{6}$ thus difference =$2\sqrt{6}\approx 5$

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$$(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$$ $$\frac{(x+2)}{(x-2)}=\frac{x^2-1}{x^2-9}$$ $$1+\frac{4}{x-2}=1+\frac{8}{x^2-9}$$ $$4x^2-36=8x-16$$ $$x^2-2x-5=0$$ Now difference of roots is $$\frac{\sqrt{D}}{a}=\sqrt{24}$$

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