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I want to show that the function $$f(x)=\frac{3-8x^{^{2}}}{1-x^{3}}$$ is one to one. For this, I suppose that $f(x)=f(y)$. So $$(x-y)(3(x^{2}+xy+y^{2})-8(x+y)-8x^{2}y^{2})=0$$ If show that $$3(x^{2}+xy+y^{2})-8(x+y)-8x^{2}y^{2})\neq0$$, then we deduce $x=y$. Please help me.

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Did you mean to have a $y$ in the denominator of $f(x)$? –  copper.hat Jan 11 '13 at 8:20
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$1-y^3$ should be $1-x^3$. Also, $8x^2 y^3$ should be $8x^2y^2$. Also, this function is not one-to-one unless you restrict the domain to, for instance, $(1,\infty)$. –  Erick Wong Jan 11 '13 at 8:28
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It isn't one to one. For example $f\left(\dfrac{\sqrt3}{\sqrt8}\right)=f\left(-\dfrac{\sqrt3}{\sqrt8}\right).$

Is one to one (with the derivative test) if its domain is restricted to $(-\infty,-1.553\ldots)$ or $(-1.553\ldots,0)$ or ...

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You can use a function grapher to plot this function. It is obvious from the graph that this is not a one-to-one function, if you do not put any restriction on the domain.

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