Question

Cho x+y=a

\(x^2+y^2=b\)

\(x^3+y^3=c\)

CM \(a^3+2c=3ab\)

Answer 1

\(a^3+2c=3ab\)

\(\Leftrightarrow\left(x+y\right)^3+2\left(x^3+y^3\right)=3\left(x+y\right)\left(x^2+y^2\right)\)

Ta có: \(VT=\left(x+y\right)^3+2\left(x^3+y^3\right)\)

\(=\left(x+y\right)^3+2\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=\left(x+y\right)\left[\left(x+y\right)^2+2\left(x^2-xy+y^2\right)\right]\)

\(=\left(x+y\right)\left(x^2+2xy+y^2+2x^2-2xy+2y^2\right)\)

\(=\left(x+y\right)\left(3x^2+3y^2\right)=3\left(x+y\right)\left(x^2+y^2\right)=VP\)

Hay ta có ĐPCM