I have a weird computations here about the (variation of the) divisor problems that involves the squarefull numbers. It is the problem to determine $\displaystyle \sum_{a^2b^3\le x} 1,$ which is answered by Richert that $$\sum_{a^2b^3\le x} 1=\zeta\left(3/2\right)x^{\frac{1}{2}}+\zeta(2/3)x^{\frac{1}{3}}+O(x^{\frac{2}{15}}).$$
Now I use the following lemmas to tackle with it.
$$\sum_{n\le x} \dfrac{1}{n^s}=\dfrac{n^{1-s}}{1-s}+\zeta(s)+O(x^{-s}), 1\not =s>0.$$
Now for the Dirichlet's convolution and for the summatory functions $F,G$ of $f,g$ are defined by $$F(x):=\sum_{n\le x} f(n), G(x):=\sum_{n\le x} g(n)$$ we have the Dirichlet's hyperbola method
$$\sum_{n\le x} f*g(n)=\sum_{n\le y} f(n)G\left(\dfrac{x}{n}\right)+\sum_{m\le x/y} g(m)F\left(\dfrac{x}{m}\right)-F\left(\dfrac{x}{y}\right)G(y).$$
Then, the last result is from the Van der Corput's method for the Bernoulli's function $B_1(t)=t-\lfloor t\rfloor-1/2$. For $R\ge 2$ be integer and an integer $N\ge 1$ with $I\subset [N,2N]$ if $f\in C^R(I,\mathbb R)$ and there exists $F>0$ such that $$\forall r\le R, x\in I, \left|f^{(r)}(x)\right|\approx FN^{-r}$$
$$\sum_{n\in I} B_1(f(n))\ll (FN^{-R})^{1/(2^{R}-1)}+F^{-1}N.$$
Now if I let $c,s$ be the characteristic function of cube and square numbers resp., and $C,S$ be their corresponding summatory functions. Then I have by the hyperbola method with $y=x^{2/5}$ that $$\sum_{a^2b^3\le x}1=\sum_{n\le x} s*c(n)=\sum_{n\le x^{2/5}} s(n)C\left(\dfrac{x}{n}\right)+\sum_{n\le x^{3/5}} c(n)S\left(\dfrac{x}{n}\right)-\lfloor x^{1/5}\rfloor^2.$$
Then, for each sum I replace $n$ by $n^2, n^3$ resp. and replace the summatory functions by $B_1$ to obtain (and assume that $x$ is the fifth power of sum integer $N$) $$\sum_{a^2b^3\le x} 1=\sum_{n\le x^{1/5}} \left\{\dfrac{x^{1/2}}{n^{3/2}}-B_1\left(\sqrt{\dfrac{x}{n}}\right)-\dfrac{1}{2}\right\}+\sum_{n\le x^{1/5}} \left\{\dfrac{x^{1/3}}{n^{2/3}}-B_1\left(\sqrt[3]{\dfrac{x}{n}}\right)-\dfrac{1}{2}\right\}-x^{2/5}\\ =x^{1/2}\sum_{n\le x^{1/5}}\dfrac{1}{n^{3/2}}+x^{1/3}\sum_{n\le x^{1/5}}\dfrac{1}{n^{2/3}}-x^{2/5}-x^{1/5}-(I_1+I_2),$$ where $I_1,I_2$ are the sum over $B_1$. By the Van der Corput's result we have $I_1,I_2\ll x^{2/15+\varepsilon}$ by dyadic division of $[1,x^{1/5}]$, and the term $x^{2/5}$ gets cancelled by the first lemma, but the term $-x^{1/5}$ from $1/2$ in each sums still occur and I don't see why is that given the answer above? Thank you for any suggestions.