Prove that $e^x, xe^x,$ and $x^2e^x$ are linearly independent over $\mathbb{R}$ 
Question: Prove that $e^x, xe^x,$ and $x^2e^x$ are linearly independent over $\mathbb{R}$.

Generally we proceed by setting up the equation
$$a_1e^x + a_2xe^x+a_3x^2e^x=0_f,$$ 
which simplifies to $$e^x(a_1+a_2x+a_3x^2)=0_f,$$ and furthermore to
$$a_1+a_2x+a_3x^2=0_f.$$
From here I think it's obvious that the only choice to make the sum the zero function is to let each scalar equal 0, but this is very weak reasoning.
As an undergraduate we learned to test for independence by determining whether the Wronskian is not identically equal to 0. But I can only use this method if the functions are solutions to the same linear homogeneous differential equation of order 3. In other words, I cannot use this method for an arbitrary set of functions. I was not given a differential equation, so I determined it on my own and got that they satisfy $$y'''-3y''+3y'-y = 0.$$
I found the Wronskian, $2e^{3x}\neq0$ for any real number. Thus the set is linearly independent. But it took me some time to find the differential equation and even longer finding the Wronskian so I'm wondering if there is a stronger way to prove this without using the Wronskian Test for Independence.

Though it is overkill and not really in the spirit of the problem (see kobe's answer for the most elementary approach), you can use the fundamental theorem of algebra (or the weaker result that a real polynomial of degree $n>1$ has at most $n$ roots) on your reduced equation to conclude that $a_3$ and $a_2$ must be zero (otherwise you'd have a positive degree polynomial with infinitely many roots). Of course $a_1$ must be zero, too: this follows just from setting $x=0$.
