From the formula $$\sum_{n \leq x}\mid \mu(n) \mid=\frac{6}{\pi^2}x(1+o(1)) $$ one can show that every sufficiently large number can be written as the sum of two square-free numbers. The number of ways to do this, say $sf_2(n)$, satisfies $$ (1-\epsilon)\frac{12-\pi^2}{\pi^2}n\leq sf_2(n) \leq (1+\epsilon)\frac{6}{\pi^2}n$$
Showing that $$sf_2(n)=O(n) \textbf{, } n \to \infty$$
I cannot replicate this procedure in the case of writing an integer as the sum of $k \ge 3$ square-free numbers e.g. a growth of $$sf_k(n)=O(n^{\alpha_k}) \textbf{, } n \to \infty$$
I hope there is a good reference for this i have skipped...