Why is $x^3-x^2+x$ injective? The function $$ f(x) = x^3-x^2+x$$ is injective (as seen on the graph), but by doing f(a) = f(b) I can't get to the point where a = b.

Can this be indicated in an analytical way (without graph) ?

Well, you should not give up hope on that $f(a)=f(b)$ approach :-):
$$f(a)=f(b)\\a^3-a^2+a=b^3-b^2+b\\a^3-b^3-(a^2-b^2)+(a-b)=0\\(a-b)(a^2+ab+b^2-a-b+1)=0$$
Which means that either $a=b$ or $a^2+ab+b^2-a-b+1=0$
Considering this as a quadratic with respect to a, we calculate it's discriminant as $$\Delta=(b-1)^2-4(b^2-b+1)=\\-3b^2+2b-3$$Which can be showed to be $<0$. 
Thus we must have $a=b$. 
