\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)(đpcm)
Có :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)^3+z^3+3z.\left(x+y\right).\left(x+y+z\right)\right]-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z.\left(x+y\right).\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3xy.\left(x+y\right)+3z.\left(x+y+z\right).\left(x+y\right)=3.\left(x+y\right).\left(xy+zx+z^2+zy\right)\)
\(=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Bài làm :
Ta có :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)^3+z^3+3z.\left(x+y\right).\left(x+y+z\right)\right]-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z.\left(x+y\right).\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3xy.\left(x+y\right)+3z.\left(x+y+z\right).\left(x+y\right)=3.\left(x+y\right).\left(xy+zx+z^2+zy\right)\)
\(=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
=> Điều phải chứng minh
