1. ## Find the solution

Find the solution in R :

2. Originally Posted by dapore
Find the solution in R :

I think you can see from the second equation that

$y = 4 - x$.

Substitute into the first equation:

$x^3 + (4 - x)^3 = 28$

$x^3 + 64 - 48x + 12x^2 - x^3 = 28$

$12x^2 - 48x + 36 = 0$

$x^2 - 4x + 3 = 0$

$(x - 1)(x - 3) = 0$

$x = 1$ or $x = 3$.

3. Originally Posted by Prove It
I think you can see from the second equation that

$y = 4 - x$.

Substitute into the first equation:

$x^3 + (4 - x)^3 = 28$

$x^3 + 64 - 48x + 12x^2 - x^3 = 28$

$12x^2 - 48x + 36 = 0$

$x^2 - 4x + 3 = 0$

$(x - 1)(x - 3) = 0$

$x = 1$ or $x = 3$.
x=1, y=4-1=3
x=3, y=4-3=1

4. Hello, dapore!

Here is another approach.
It's no shorter, but it's rather "cute".

Find the solution in R: . $\begin{array}{cccc}x^3 + y^3 &=& 28 & [1] \\ x + y &=& 4 & [2] \end{array}$

$\text{Cube [2]: }\;(x+y)^3 \:=\:4^3 \quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:64$

$\text{and we have: }\;\underbrace{x^3 + y^3}_{\text{This is 28}} + 3xy\underbrace{(x+y)}_{\text{This is 4}} \:=\:64 \quad\Rightarrow\quad 28 + 3xy(4) \:=\:64$

. . . . . . . . . $12xy \:=\:36 \quad\Rightarrow\quad y \:=\:\tfrac{3}{x}\;\;[3]$

Substitute into [2]: . $x + \tfrac{3}{x} \:=\:4 \quad\Rightarrow\quad x^2 - 4x + 3 \:=\:0$

. . . . . . $(x-1)(x-3) \:=\:0 \quad\Rightarrow\quad x \:=\:1,\:3$

Substitute into [3]: . $y \:=\:3,\:1$

Solutions: . $(x,y) \:=\1,3),\3,1)" alt="(x,y) \:=\1,3),\3,1)" />

5. Thank you my friends

6. $x^{3}+y^{3}=(x+y)\left( (x+y)^{2}-3xy \right)=4(16-3xy)=28,$ thus $xy=3.$ (1)

Now put the second equation into (1) and get $4x-x^2=3\implies x^2-4x+3=(x-1)(x-3)=0.$

The solutions are $(x,y)=(1,3)$ and $(x,y)=(3,1).$