Áp dụng hàm đẳng thức vào biểu thức trên ta được:

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

=(x^3-3.x^2.y+3.x.y^2-y^3)+(y^3-3.y^2.z+3.y.z^2-z^3)+(z^3-3.z^2.x+3.z.x^2-x^3)

=-3.x^2.y+3.x.y^2-3.y^2.z+3.y.z^2-3.z^2.x+3.z.x^2

=3.(.x^2.y+x.y^2-y^2.z+y.z^2-z^2.x+z.x^2)..

đén đây thì mình chịu, mong bạn thông cảm cho mình nha!(~~__~~)

ai giúp hộ kìa

Đặt \(y+z=p\)

Khi đó \(M=\left(x+p\right)^3+\left(x-p\right)^3\)\(=x^3+3x^2p+3xp^2+p^3+x^3-3x^2p+3xp^2-p^3\)\(=2x^3+6xp^2=2x^3+6x\left(y+z\right)^2=N\) (vì \(y+z=p\))

 Từ đó ta có đpcm.

Ta có

\(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy\left(x+y\right)\)

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

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

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

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

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

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

pls, mai mk phải nộp bài r

pls, mai mk phải nộp bài r

pls, mai mk phải nộp bài r

x^3+y^3+z^3=3xyz

<=>x^3+y^3+z^3-3xyz=0

<=>(x+y+z).(x^2+y^2+z^2-xy-yz-zx)=0

<=>x^2+y^2+z^2-xy-yz-zx=0 (vì x,y,z > 0 nên x+y+z > 0)

<=>2x^2+2y^2+2z^2-2xy-2yz-2zx=0

<=>(x-y)^2+(y-z)^2+(z-x)^2=0

<=>x-y=0;y-z=0;z-x=0

<=>x=y=z (ĐPCM)

k mk nha

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

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

\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)

           ~ Chúc bạn học tốt~

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

\(=x^3+3x^2yz+3xy^2z+3xyz^2+y^3+z^3-x^3-y^3-z^3\)

\(=3x^2yz+3xy^2z+3xyz^2\)

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

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

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

\(=3\times\left(x+y\right)\times\left(y+z\right)\times\left(x+z\right)\)

Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3

= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3

Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0

=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3

=3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )

= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )

\((x^2+y^2)^3+(z^2-x^2)^3-(y^2+z^2)^3\)

\(=-3[x^4y^2-x^4z^2-x^2y^2z^2+x^2z^4-x^2y^4+x^2y^2z^2+y^4z^2-y^2z^4\)

\(=-3[x^2(x^2y^2-x^2z^2-z^2y^2+z^4)-y^2(x^2y^2-x^2z^2-z^2y^2+z^4)\)

\(=-3(x^2-y^2)(x^2y^2-x^2z^2-z^2y^2+z^4)\)

\(=-3(x^2-y^2[x^2(y^2-z^2)-z^2(y^2-z^2)]\)

\(=-3(x^2-y^2)(x^2-z^2)(y^2-z^2)\)

\(=-3(x-y)(x+y)(x-z)(x+z)(y+z)(y-z)\)