Question

[Nguyễn Văn Thố]

rút gọn biểu thức (a+b)^3 +(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)

Answer 1

[Lightning Farron]

Đặt \(x=a+b;y=b+c;z=c+a\) ta có:

\(x^3+y^3+z^3-3xyz\)

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

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

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

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

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

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

Thay vào ta có:\(\left(a+b+b+c+c+a\right)\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-\left(a+b\right)\left(b+c\right)-\left(b+c\right)\left(c+a\right)-\left(c+a\right)\left(a+b\right)\right]\)

\(=\left(2a+2b+2c\right)\left(a^2-ab-ac+b^2-bc+c^2\right)\)

\(=2\left(a+b+c\right)\left(a^2-ab-ac+b^2-bc+c^2\right)\)

Answer 2

[Nguyễn Văn Thố]

cảm ơn bạn nhìu