Đáp án:
Giải thích các bước giải:
1) `(x-y)^3+(y-z)^3+(z-x)^3`
`=x^3-y^3-3xy(x-y)+y^3-z^3-3yz(y-z)+z^3-x^3-3yz(z-x)`
`=-3xy(x-y)-3yz(y-z)-3zx(z-x)`
`=-3y((x^2-xy)+yz-z^2))-3zx(z-x)`
`=-3y(x^2-z^2-xy+yz)-3zx(z-x)`
`=-3y((x-z)(x+z)-y(x-z))+3zx(x-z)`
`=-3(x-z)((y(x+z)-y^2)-zx)`
`=-3(x-z)(xy+yz-y^2-zx)`
`=-3(x-z)(xy-y^2-zx+yz)`
`=-3(x-z)(y(x-y)-z(x-y))`
`=-3(x-z)(x-y)(y-z)`
`=3(z-x)(x-y)(y-z)`
`=3(z-x)(x-y)(y-z)`
2) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3x^2y+3xy^2+3x^2z+3xz^2+3y^2z+3yz^2+6xyz\)\(=3\left(xyz+x^2y+x^2z+xz^2+xy^2+y^2z+xyz+yz^2\right)\)
\(=3\left[xy\left(x+z\right)+xz\left(x+z\right)+y^2\left(x+z\right)+yz\left(x+z\right)\right]\)
\(=3\left(x+z\right)\left[xy+xz+y^2+yz\right]\)
\(=3\left(x+z\right)\left[x\left(y+z\right)+y\left(y+z\right)\right]\)
\(=3\left(x+z\right)\left(y+z\right)\left(x+y\right)\)
3) Đặt \(x+y-z=a, x+z-y=b, y+z-x=c\) thì \(a+b+c=x+y+z\)
Đẳng thức cần chứng minh trở thành \((a+b+c)^3-a^3-b^3-c^2=3(a+b)(b+c)(c+a)\)
Ta có \((a+b+c)^3=[(a+b)+c]^3=(a+b)^3+c^3+3(a+b)c(a+b+c)\)
\(=a^3+b^3+c^3+3ab(a+b)+3(a+b)c(a+b+c)\)
\(=a^3+b^3+c^3+3(a+b)[ab+c(a+b+c)]\)
\(=a^3+b^3+c^3+3(a+b)[a(b+c)+c(b+c)]\)
\(=a^3+b^3+c^3+3(a+b)(b+c)(c+a)\)
Thay a, b, c bằng các biến đã đặt rồi rút gọn ta được kết quả là `24xyz`
