a) Theo giả thiết, ta có: \(x+y=1\Rightarrow\left(x+y\right)^3=1^3\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\Leftrightarrow x^3+y^3+3xy=1\)
b) Theo giả thiết, ta có: \(x+y=5\Rightarrow\left(x+y\right)^2=5^2\Leftrightarrow x^2+2xy+y^2=25\Leftrightarrow x^2+y^2=19\)
Lại có: \(x+y=5\Rightarrow\left(x+y\right)^3=5^3\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=125\Leftrightarrow x^3+y^3=80\)
Ta có:
- \(x^2+y^2=19\Rightarrow\left(x^2+y^2\right)^2=19^2\Leftrightarrow x^4+2x^2y^2+y^4=361\Leftrightarrow x^4+y^4=343\)
- \(x^5+y^5=\left(x+y\right)\left(x^4+y^4\right)-xy\left(x^3+y^3\right)=5.343-3.80=1475\)
c) Theo giả thiết, ta có: \(x+y=3\Rightarrow\left(x+y\right)^2=3^2\Leftrightarrow x^2+2xy+y^2=9\Leftrightarrow2xy=4\Leftrightarrow xy=2\)
Do đó: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2\right)-xy\left(x+y\right)=3.5-2.3=9\)
