$\Leftrightarrow9x^6-51x^5+201x^4-350x^3+219x^2+9x-37=27x^6-54x^5+9x^4+28x^3-3x^2-6x-1$ $\Leftrightarrow-18x^6+3x^5+192x^4-378x^3+222x^2+15x-36=0$ $\Leftrightarrow-3(x-1)^3(x+4)(2x-3)(3x+1)=0$ after IVT and RRT So $x\in\boxed{\left\{-4,-\frac13,1,\frac32\right\}}$.

If $x^2+3x-4=a,2x^2-5x+3=b$ then $3x^2-2x-1=a+b$ So we have $a^3+b^3=(a+b)^3$ or $3ab(a+b)=0$ So we need to solve $x^2+3x-4=0 \to x=1,-4$; $2x^2-5x+3=0 \to x=1,\frac{3}{2}$; $3x^2-2x-1=0 \to x=1,-\frac{1}{3}$ And total $x\in\boxed{\left\{-4,-\frac13,1,\frac32\right\}}$

jasperE3 wrote: after IVT and RRT. what is IVT and RRT?

Intermediate value theorem and rational root theorem

$$\left ( x^2+3x-4 \right )^3+\left ( 2x^2-5x+3 \right )^3=\left ( 3x^2-2x-1 \right )^3$$$$\left ( x-1 \right )^3\left ( x+4 \right )^3+\left ( x-1 \right )^3\left ( 2x-3 \right )^3-\left ( x-1 \right )^3\left ( 3x+1 \right )^3=0$$$$\left ( x-1 \right )^3\left ( \left ( x+4 \right )^3+\left ( 2x-3 \right )^3-\left ( 3x+1 \right )^3 \right )=0$$$$\left ( x-1 \right )^3\left ( \left ( x+4 \right )^3-\left ( x+4 \right )\left ( 19x^2-13x+7 \right ) \right )=0$$$$\left ( x-1 \right )^3\left ( x+4 \right )\left ( 21x-18x^2+9 \right )=0$$$$x=\left \{ -4;-\frac{1}{3};1;\frac{3}{2} \right \}$$

$$\left(x^2+3x-4\right)^3+\left(2x^2-5x+3\right)^3=\left(3x^2-2x-1\right)^3$$$$\iff -18 x^6 + 3 x^5 + 192 x^4 - 378 x^3 + 222 x^2 + 15 x - 36 = 0$$$$\iff -3 (x - 1)^3 (x + 4) (2 x - 3) (3 x + 1) = 0$$$$\iff x\in \left \{-4,-\frac{1}{3},1,\frac{3}{2} \right \}$$

ZETA_in_olympiad wrote: $$\left(x^2+3x-4\right)^3+\left(2x^2-5x+3\right)^3=\left(3x^2-2x-1\right)^3$$$$\iff -18 x^6 + 3 x^5 + 192 x^4 - 378 x^3 + 222 x^2 + 15 x - 36 = 0$$$$\iff -3 (x - 1)^3 (x + 4) (2 x - 3) (3 x + 1) = 0$$$$\iff x\in \left \{-4,-\frac{1}{3},1,\frac{3}{2} \right \}$$ Why write a solution that has already happened?

Leviathan_ wrote: ZETA_in_olympiad wrote: $$\left(x^2+3x-4\right)^3+\left(2x^2-5x+3\right)^3=\left(3x^2-2x-1\right)^3$$$$\iff -18 x^6 + 3 x^5 + 192 x^4 - 378 x^3 + 222 x^2 + 15 x - 36 = 0$$$$\iff -3 (x - 1)^3 (x + 4) (2 x - 3) (3 x + 1) = 0$$$$\iff x\in \left \{-4,-\frac{1}{3},1,\frac{3}{2} \right \}$$ Why write a solution that has already happened? There is nothing wrong with doing that. It is obviously not a direct copy paste so it's fine.