My knowledge of algebra is still insufficient to solve this problem. Any help in solving the system of equations would be greatly appreciated.
$$ xy(x+y)=30\\ x^3+y^3=35 $$
My knowledge of algebra is still insufficient to solve this problem. Any help in solving the system of equations would be greatly appreciated.
$$ xy(x+y)=30\\ x^3+y^3=35 $$
$(x+y)^3=x^3+y^3+3xy(x+y)$
$\Rightarrow (x+y)^3=35+90=125\Rightarrow (x+y)=5$
$\Rightarrow xy(x+y)=5xy=30\Rightarrow xy=6$
$(x-y)^2=(x+y)^2-4xy=25-24=1\Rightarrow (x-y)=\pm1$
case 1: $x-y=1$ and $x+y=5$ ,$\Rightarrow x=3,y=2$
case 2: $x-y=-1$ and $x+y=5$, $\Rightarrow x=2,y=3$
Note that $$ (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 = x^3+y^3 + 3xy(x+y) = 35 + 3 \cdot 30 = 125 $$ and therefore $x+y = \sqrt[3]{125} = 5$ and $xy = \frac{30}{x+y} = 6$.
Thus $y = 5-x$ and $6 = xy = x(5-x)$, and so $x^2-5x+6=0$, which implies $x \in \{2,3\}$ and so $y \in \{3,2\}$. Thus, either $(x,y)=(2,3)$ or $(x,y) = (3,2)$...
Distribute the first equation and add up that one with the second.
You get $${}(X + y)^3 = 125{}$$
Solve from there