Algebra exercise asked by the teacher Imad zak in the math group

Exercise:

If $x+y=1\,\,\,\,\And \,\,\,{{x}^{2}}+{{y}^{2}}=2$ , Evaluate ${{x}^{3}}+{{y}^{3}}$

Solution: we know that ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}$

So ${{x}^{3}}+{{y}^{3}}+3\left( {{x}^{2}}y+x{{y}^{2}} \right)=1\Leftrightarrow {{x}^{3}}+{{y}^{3}}=1-3\left( {{x}^{2}}y+x{{y}^{2}} \right)$

But ${{x}^{2}}=2-{{y}^{2}}\Leftrightarrow {{x}^{3}}+{{y}^{3}}=1-3\left( \left( 2-{{y}^{2}} \right)y+x{{y}^{2}} \right)$

$\Leftrightarrow {{x}^{3}}+{{y}^{3}}=1-3\left( 2y-{{y}^{3}} \right)-3x\left( 2-{{x}^{2}} \right)=1-6y+3{{y}^{3}}-6x+3{{x}^{3}}$

$\Leftrightarrow {{x}^{3}}+{{y}^{3}}=1-6\left( y+x \right)+3\left( {{x}^{3}}+{{y}^{3}} \right)\Leftrightarrow -2\left( {{x}^{3}}+{{y}^{3}} \right)=1-6=-5\Leftrightarrow {{x}^{3}}+{{y}^{3}}=\frac{5}{2}$