Does $\Delta u = f(|x|)$ mean that $u=v(|x|)$ for some $v$? If $u$ solves $\Delta u = f(|x|)$ in $B_1 \subset \mathbb{R}^n$ for some continuous function $f: [0,\infty) \to \mathbb{R}$ does that mean that $u(x) = v(|x|)$ for some function $v$?

I think this is true result about radially symmetric solutions, but I may need need one more condition that $u = 0$ on $\partial B_1$. Anyway, I am having a difficult time showing this result.
Hint: This might not be true 

Take $$u(x) = \sum_{i=1}^{n}x_i \implies \Delta u  = 0 $$

or 
 $$u(x) =\sum_{i=1}^{n}x_i^4\implies \Delta u  = 12|x|^2$$
 $$u(x) =\sum_{i=1}^{n}(x_i^4 +x_i)\implies \Delta u  = 12|x|^2 $$
Hint What happens if you take $u : \Bbb R^n \to \Bbb R$ to be any nonradial harmonic function (e.g., the projection onto the first component)?