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# If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b)

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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]
sa800 wrote:
my question is why or where are people getting -1,0,1 from? I assume they factor …. but then Why do people automatically set the quadratic equation equal to zero after factoring?

nowhere on the question does it say that….(x)(x+1)(x-1) = 0…..

is there a fundamental theorem im missing?

You are missing the crucial piece of information given in the stem: $$x^3-x=(x-a)(x-b)(x-c)$$ for ALL numbers $$x$$.

Since $$x^3-x=(x-a)(x-b)(x-c)$$ is true for ALL $$x-es$$ than it must also be true for $$x=0$$, $$x=1$$ and $$x=-1$$. Setting x to 0 1 and -1 allows us to determine the values of a, b, and c.

Hope it helps.
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If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]