Question

Cho \(x+y=a,x^2+y^2=b,x^3+y^3=c\). Tính giá trị biểu thức: \(a^3-3ab+2c\)

Answer 1

Ta có :

\(a^3-3ab+2c=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)

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

\(=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x+y\right)^3-6xy\left(x+y\right)\)

\(=3\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)-6xy\left(x+y\right)\)

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

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

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

\(=0\)