Sistema

{(2·x - 1)^2 = y + (4·x + 1)·(x - 2)

{x^3 + y = (x - 2)^3 + 6·x^2

----------------------------------

{4·x^2 - 4·x + 1 = y + (4·x^2 - 7·x - 2)

{x^3 + y = (x^3 - 6·x^2 + 12·x - 8) + 6·x^2

--------------------------------

{- 4·x + 1 = y + (- 7·x - 2)

{y = (- 6·x^2 + 12·x - 8) + 6·x^2

------------------------------

dalla seconda:

y = 12·x - 8

procedo quindi per sostituzione:

- 4·x + 1 = (12·x - 8) + (- 7·x - 2)

- 4·x + 1 = 5·x - 10

5·x + 4·x = 11

9·x = 11----> x = 11/9

y = 12·(11/9) - 8----> y = 20/3

soluzione sistema: [x = 11/9 ∧ y = 20/3]

(2x - 1)^2 = y + (4x + 1)(x - 2);  (1)

x^3  + y = (x - 2)^3 + 6x^2;  (2)

4x^2 + 1 - 4x = y + 4x^2 - 8x + x - 2;  (1)

x^3  + y = x^3 - 6x^2 + 12x - 8 + 6x^2;  (2)     ( i termini di 2° grado e 3° grado si annullano);

- 4x + 8x - x - y = - 2 - 1;  (1)

y - 12x = - 8;  (2)

3x - y = - 3;  (1)

y = 12x - 8;  (2)  sostituiamo  la (2) nella  (1);

3x - (12x - 8) = - 3;  (1)

3x - 12x  + 8 = - 3;

- 9x = - 11;

x = 11/9;

y = 12x - 8;

y = 12 * (11/9) - 8 = 4 * (11/3) - 8;

y = 44/3 - 24/3 = 20/3.

Ciao @alby

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$\small \begin{Bmatrix}
(2x-1)^2&=&y+(4x+1)(x-2)\\
x^3+y&=&(x-2)^3+6x^2\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
4x^2-4x+1&=&y+4x^2-8x+x-2\\
x^3+y&=&x^3\cancel{-6x^2}+12x-8\cancel{+6x^2}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
4x^2-4x+1&=&y+4x^2-7x-2\\
x^3+y&=&x^3+12x-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\cancel{4x^2}-4x-y\cancel{-4x^2}+7x&=&-2-1\\
\cancel{x^3}+y\cancel{-x^3}-12x&=&-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
3x-y&=&-3\\
y-12x&=&-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
3x-(12x-8)&=&-3\\
y&=&12x-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
3x-12x+8&=&-3\\
y&=&12x-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-9x&=&-3-8\\
y&=&12x-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
-9x&=&-11\\
y&=&12x-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
\dfrac{\cancel9x}{\cancel9}&=&\dfrac{11}{9}\\
y&=&12x-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{11}{9}\\
y&=&\cancel{12}^4·\dfrac{11}{\cancel9_3}-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{11}{9}\\
y&=&4·\dfrac{11}{3}-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{11}{9}\\
y&=&\dfrac{44}{3}-8\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{11}{9}\\
y&=&\dfrac{44-24}{3}\\
\end{Bmatrix}$

$\small \begin{Bmatrix}
x&=&\dfrac{11}{9}\\
y&=&\dfrac{20}{3}\\
\end{Bmatrix}$