\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+2xy+2xz+2yz-x^3-y^3-z^3\)
\(=2xy+2xz+2yz\)
\(=2\left(xy+xz+yz\right)\)
Đc chưa ?
Phương Đỗ Sai rùi bạn.
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y+z\right)\left(x+y\right)z+z^3-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+3\left(x+y+z\right)\left(x+y\right)z+z^3-x^3-y^3-z^3\)
\(=3xy\left(x+y\right)+3\left(x+y+z\right)\left(x+y\right)z\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)\(=3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)\)
\(=3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[\left(xy+xz\right)+\left(yz+z^2\right)\right]\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
P/s: Bài viết sử dụng 1 công thức đặc biệt: (a+b)^3 = a^3 + b^3 + 3ab(a+b)
Công thức được suy ra từ hàng đẳng thức số 4 khi:
(a+b)^3 = a^3 + 3*a^2*b + 3*a*b^2 + b^3 = a^3 + 3*a*b*(a+b) + b^3
