Question

Cho \(x+y+z=0\)

Tính giá trị biểu thức \(B=\dfrac{x^3+y^3+z^3}{3xyz}\)

Answer 1

Ta có:

\(B=\dfrac{x^3+y^3+z^3}{3xyz}\)

\(=\dfrac{x^3+y^3+z^3-3xyz}{3xyz}+1\)

\(=\dfrac{\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz}{3xyz}+1\)

\(=\dfrac{\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)}{3xyz}+1\)\(=\dfrac{0-0}{3xyz}+1=1\)