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The function is $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}$ and they are asking us to find out what $f(-x)$ is?

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  • $\begingroup$ Why don't you write it as $f(x-1/x) = (x-1/x)^3 + 3(x-1/x)$? $\endgroup$ – Parth Kohli Jul 8 '16 at 18:16
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Observe that $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}=(x-\frac{1}{x})^3+3\cdot x\cdot \frac{1}{x}\cdot(x-\frac{1}{x})=(x-\frac{1}{x})^3+3(x-\frac{1}{x})$

Hence, we can say that $$f(z)=z^3+3z$$ where $z$ is any real variable.

So we have that $$\color{red}{f(-x)}=\color{blue}{-x^3-3x}$$

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A slightly different approach from the others:

Note that every real number $x$ can be written in the form $a-\frac1a$ for some $a\ne0$. Then $-x=-\left(a-\frac1a\right)=(-a)-\frac1{-a}$, so $$f(-x)=(-a)^3-\frac1{(-a)^3}=-\left(a^3-\frac1{a^3}\right)=-f(x).$$

If you need an explicit expression for $f(-x)$, then either solve for $a$ and substitute, or use one of the methods described in the other answers.

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A general method for solving equations of the type:

$$f(g(x))=z(x)$$

Let:

$$u=x-\frac{1}{x}$$

Here we let $u$ be equal to $g(x)$.

$$ux=x^2-1$$

$$x^2-ux-1=0$$

Using the quadratic formula we have:

$$x=\frac{u \pm \sqrt{u^2+4}}{2}$$

Note in a way we found an inverse for $x-\frac{1}{x}$ even though it does not have an inverse function. Now we have:

$$f(u)=(\frac{u \pm \sqrt{u^2+4}}{2})^3-\frac{1}{(\frac{u \pm \sqrt{u^2+4}}{2})^3}$$

And:

$$f(x)=(\frac{x \pm \sqrt{x^2+4}}{2})^3-\frac{1}{(\frac{x \pm \sqrt{x^2+4}}{2})^3}$$

Which can be simplified down to get the result $f(x)=x^3+3x$.

$$f(-x)=(\frac{-x \pm \sqrt{x^2+4}}{2})^3-\frac{1}{(\frac{-x \pm \sqrt{x^2+4}}{2})^3}$$

Again this can be simplified to get your desired result $f(-x)=-x^3-3x$.

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  • $\begingroup$ Wolfram alpha confirms this gives $f(x)=x^3+3x$, see this wolframalpha.com/input/…. and this wolframalpha.com/input/…. For some reason I had to put in in this form because the other forms where not working. $\endgroup$ – Ahmed S. Attaalla Jul 8 '16 at 20:21
  • $\begingroup$ Thanks. Why does Wolfram give this warning : "An attempt was made to fix mismatched parentheses, brackets, or braces." and why it won't show step by step simplification of that expression? Also, when I inputed "Simplify f(x+1\x)=x^3+1\x^3" it did not gave me -x^3-3x form? $\endgroup$ – Marko Savic Jul 8 '16 at 21:04
  • $\begingroup$ No problem. Sometimes it just does that because the expression is to much, but I could be wrong. And I'm not sure about the other thing, it could be it interpreted wrong. $\endgroup$ – Ahmed S. Attaalla Jul 8 '16 at 21:07
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write it as follows $$f\left( x-\frac { 1 }{ x } \right) =x^{ 3 }-\frac { 1 }{ x^{ 3 } } ={ \left( x-\frac { 1 }{ x } \right) }^{ 3 }+3\left( x-\frac { 1 }{ x } \right) $$ it means our function is $$f\left( x \right) ={ x }^{ 3 }+3x$$ now

$$f\left( -x \right) =-{ x }^{ 3 }-3x$$

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As per Parth Kohli's comment we can write $$f\left(x - \frac{1}{x}\right) = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right).$$

That is, $f(x) = x^3 + 3x$ so $f(-x) = (-x)^3 + 3(-x) = -(x^3 + 3x)$.

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