Find the solution in R :
I think you can see from the second equation that
$\displaystyle y = 4 - x$.
Substitute into the first equation:
$\displaystyle x^3 + (4 - x)^3 = 28$
$\displaystyle x^3 + 64 - 48x + 12x^2 - x^3 = 28$
$\displaystyle 12x^2 - 48x + 36 = 0$
$\displaystyle x^2 - 4x + 3 = 0$
$\displaystyle (x - 1)(x - 3) = 0$
$\displaystyle x = 1$ or $\displaystyle x = 3$.
Hello, dapore!
Here is another approach.
It's no shorter, but it's rather "cute".
Find the solution in R: .$\displaystyle \begin{array}{cccc}x^3 + y^3 &=& 28 & [1] \\ x + y &=& 4 & [2] \end{array}$
$\displaystyle \text{Cube [2]: }\;(x+y)^3 \:=\:4^3 \quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:64$
$\displaystyle \text{and we have: }\;\underbrace{x^3 + y^3}_{\text{This is 28}} + 3xy\underbrace{(x+y)}_{\text{This is 4}} \:=\:64 \quad\Rightarrow\quad 28 + 3xy(4) \:=\:64$
. . . . . . . . . $\displaystyle 12xy \:=\:36 \quad\Rightarrow\quad y \:=\:\tfrac{3}{x}\;\;[3]$
Substitute into [2]: .$\displaystyle x + \tfrac{3}{x} \:=\:4 \quad\Rightarrow\quad x^2 - 4x + 3 \:=\:0$
. . . . . . $\displaystyle (x-1)(x-3) \:=\:0 \quad\Rightarrow\quad x \:=\:1,\:3$
Substitute into [3]: .$\displaystyle y \:=\:3,\:1$
Solutions: .$\displaystyle (x,y) \:=\1,3),\3,1)$
$\displaystyle x^{3}+y^{3}=(x+y)\left( (x+y)^{2}-3xy \right)=4(16-3xy)=28,$ thus $\displaystyle xy=3.$ (1)
Now put the second equation into (1) and get $\displaystyle 4x-x^2=3\implies x^2-4x+3=(x-1)(x-3)=0.$
The solutions are $\displaystyle (x,y)=(1,3)$ and $\displaystyle (x,y)=(3,1).$