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$\left(2x-5\right)^3+\left(2-x\right)^3=\left(x-3\right)^3+\left(x-3\right)\left(7x^2-2052x+2054\right)$

$\left(2x-5\right)^3+\left(2-x\right)^3-\left(x-3\right)^3-\left(x-3\right)\left(7x^2-2052x+2054\right)=0$

$\left(2x-5+2-x\right)\left\lbrack\left(2x-5\right)^2-\left(2x-5\right)\left(2-x\right)+\left(2-x\right)^2\right\rbrack-\left(x-3\right)\left\lbrack\left(x-3\right)^2+7x^2-2052x+2054\right\rbrack=0$

$\left(x-3\right)\left(4x^2-20x+25+2x^2-9x+10+4-4x+x^2\right)-\left(x-3\right)\left(x^2-6x+9+7x^2-2052x+2054\right)=0$

$\left(x-3\right)\left(7x^2-33x+39\right)-\left(x-3\right)\left(8x^2-2058x+2063\right)=0$

$\left(x-3\right)\left(-x^2+2025x-2024\right)=0$

$\left(x-3\right)\left(x^2-2025x+2024\right)=0$

$\left(x-3\right)\left(x-1\right)\left(x-2024\right)=0$

$\left[\begin{matrix}x-3=0 \\ x-1=0 \\ x-2024=0\end{matrix}\right.$

$\left[\begin{matrix}x=3 \\ x=1 \\ x=2024\end{matrix}\right.$

Vậy $x\in\left\lbrace1;3;2024\right\rbrace$.