$\text{1) x³ + y³=(x+y)(x²-xy+y²)}$
$\text{Ta có x+y=2}$
$\text{⇒ (x+y)²=4}$
$\text{⇔ x²+2xy+y²=4}$
$\text{hay x²+y²+2.(-6)=4}$
$\text{x²+y²-12=4}$
$\text{⇒ x²+y²=16}$
$\text{Thay x+y=2;xy=-6;x²+y²=16 vào (x+y)(x²-xy+y²) ta được }$
$\text{2.[16-(-6)]=2.22=44}$
$\text{2/xy=-6⇒x²y²=36}$
$\text{x²+y²=16 (chứng minh trên)}$
$\text{⇒ (x²+y²)²=256}$
$\text{⇔ $x^{4}$ +$y^{4}$+2x²y²=256}$
$\text{Thay x²y²=36 vào $x^{4}$ +$y^{4}$+2x²y²=256 ta được }$
$\text{$x^{4}$ +$y^{4}$+2.36=256}$
$\text{⇔ $x^{4}$ +$y^{4}$+72=256}$
$\text{⇔ $x^{4}$ +$y^{4}$=184}$
