$x{f}^{\text{'}}(2x-1)-f(2x-1)=x^3\left(1\right)$. Đặt $t=2x-1\iff x=\frac{t+1}{2}$, đẳng thức (1) thành

$\frac{\text{t}+1}{2}{\text{f}}^{\text{'}}\left(\text{t}\right)-\text{f}\left(\text{t}\right)={\left(\frac{\text{t}+1}{2}\right)}^3\iff \frac{1}{{(\text{t}+1)}^2}{\text{f}}^{\text{'}}\left(\text{t}\right)-\frac{2}{{(\text{t}+1)}^3}\text{f}\left(\text{t}\right)=\frac{1}{4}\iff {\left(\frac{1}{{(\text{t}+1)}^2}\text{f}\left(\text{t}\right)\right)}^{\text{'}}=\frac{1}{4}$

Như vậy ta có ${\left(\frac{1}{{(x+1)}^2}f\left(x\right)\right)}^{\text{'}}=\frac{1}{4}$. (2)

Lấy nguyên hàm hai vế đẳng thức (2) ta được

$\int {\left(\frac{1}{{(x+1)}^2}f\left(x\right)\right)}^{\text{'}}dx=\int \frac{1}{4}dx\iff \frac{1}{{(x+1)}^2}f\left(x\right)=\frac{1}{4}x+C$

Lại có f(1) = 1 => C = 0. Do đó $f\left(x\right)=\frac{x{(x+1)}^2}{4}$