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Let $ f: \mathbb{R} \mapsto \mathbb{R} $ be a continuous function. Prove that there is a $ c $ such that $ f(c)=c^3 $.

But the functions $ f(x)=(x-1)^3 $ or $ f(x)=(x-a)^3, a>0 $ etc, don't satisfy the statement.

Am I right ?

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    yes you are, maybe you forgot a hypothesis on $f$. – 2017-02-15
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    Another counterexample : $x\mapsto x^3+a$ where $a\neq0$. – 2017-02-15
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    An equivalent statement: the graph of every continous function $f:\Bbb R\to\Bbb R$ intersects the curve $y=x^3$. This is obvoiusly false. – 2017-02-15
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    @ajotatxe This is an excellent way to think about this problem. You might consider posting that as an answer. – 2017-02-15
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    @user3798897 Thank you, but I think that my comment does not answer OP's question. OP simply wants to know if his/her counterexamples are valid. – 2017-02-15

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I will exam $f(x)=(x-1)^3$.

If there is a c such that $f(c)=c^3$, then $c^3=(c-1)^3$. So $c^3=c^3-c^2+c-1$ or $-c^2+c-1=0$. Since $\delta = 1^2-4*(-1)*(-1)<0$, this equation has no real roots.

It means that you are correct about the $f(x)=(x-1)^3$.

The others, you can exam by the same method.