nice exercise asked by Imad Zak using sum and product

Exercise:

Solve in $\mathbb{R}$ , ${{x}^{3}}+{{y}^{3}}=5\,\,,\,\,\,{{x}^{2}}y+x{{y}^{2}}=1$

We have ${{x}^{2}}y+x{{y}^{2}}=1\Leftrightarrow xy\left( x+y \right)=1\Leftrightarrow PS=1$

Also we know that ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$

So ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}}+2xy-2xy-xy \right)=\left( x+y \right)\left( {{\left( x+y \right)}^{2}}-3xy \right)=5$

i.e $PS=1\,\,\And \,\,\,\,S\left( {{S}^{2}}-3P \right)=5$

so $S\left( {{S}^{2}}-3P \right)=5\Leftrightarrow {{S}^{3}}-3SP=5\Leftrightarrow {{S}^{3}}-3=5\Leftrightarrow {{S}^{3}}=8\Leftrightarrow S=2$  and $P=\frac{1}{2}$

now solving the equation ${{x}^{2}}-Sx+P=0\Leftrightarrow {{x}^{2}}-2x+\frac{1}{2}=0$

${{x}^{2}}-2x+1-1+\frac{1}{2}=0\Leftrightarrow {{\left( x-1 \right)}^{2}}=\frac{1}{2}\Leftrightarrow x=1\pm \frac{1}{\sqrt{2}}=\frac{2\pm \sqrt{2}}{2}$ and $y=\frac{2\pm \sqrt{2}}{2}$