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.

Suppose $a_1e^x + a_2xe^x+a_3x^2e^x=0 $ for all $x$.
Setting $x=0$ shows that $a_1 = 0$.
Now note that $a_2xe^x+a_3x^2e^x=0 $ for all $x$ and hence
$a_2e^x+a_3xe^x=0 $ for all $x \neq 0$. Taking limits as $x \to 0$ shows
that $a_2 = 0$, and setting $x=1$ shows that $a_3 = 0$.
