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) ?

Let $f(a) = f(b)$, then $f(a)-f(b) = 0 \iff a^3-a^2+a -b^3+b^2-b = 0 \iff a-b = a^2(1-a) -b^2(1+b)$
But this means
$2(a-b) = a^2(1-a) -b^2(1+b) + a - b = (f(a)-f(b)) = 0$, from which $a = b$ follows. 
