Ta có: \(\left(b-c\right)^3+\left(c-a\right)^3-\left(a-b\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(b-c+c-a\right)\left[\left(b-c\right)^2-\left(b-c\right)\left(c-a\right)+\left(c-a\right)^2\right]-\left(a-b\right)\left[1+3\left(b-c\right)\left(c-a\right)\right]\)

\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2\right)-\left(a-b\right)\left(1+3bc-3ab-3c^2+3ac\right)\)

\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2+1+3bc-3ab-3c^2+3ac\right)\)

\(=\left(b-a\right)\left(b^2-2ab+a^2+1\right)\)

\(=\left(b-a\right)^3+\left(b-a\right)\)

\(=b^3-3b^2a+3ba^2-a^3+b-a\)

M=a^3+b^3+c^3-3abc/(a-b)^3+(b-c)^3+(c-a)^3

nè ban

\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Đặt a+b=x ; b+c=y; c+a=z ta có:

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

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

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

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

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

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

xong thay vào

b) F=3.(b-c)-(a+c)+5.(a-b-c)

   F=3b-3c-a-c+5a-5b-5c

   F=(-a+5a)+(3b-5b)+(-3c-c-5c)

  F= 4a+(-2b)+(-9c)

  F=4a-2b-9c

a) E=2.(a+b)+3.(a-c)-4(b+c)

   E=2a+2b+3a-3c-4b-4c

   E=(2a+3a)+(2b-4b)+(-3c-4c)

   E=5a+(-2b)+(-7c)=5a-2b-7c

c)G=7.(a-b-c)+4.(b+c)-6(b-c-a)

G=7a-7b-7c+4b+4c-6b+6c+6a

G=(7a+6a)+(-7b+4b-6b)+(-7c+4c+6c)

G=13a+(-9b)+3c

G=13a-9b+3c

Áp dụng hằng đẳng thức dưới dạng 

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

\(\left(a+b+c\right)^3+\left(a-b-c\right)^3=\left(2a\right)^3-3\left(a+b+c\right)\left(a-b-c\right).2a\)

\(\left(b-c-a\right)^3+\left(c-a-b\right)^3=\left(-2a\right)^3-3\left(b-c-a\right)\left(c-a-b\right).\left(-2a\right)\)

\(\Rightarrow\left(a+b+c\right)^3+\left(a-b-c\right)^3+\left(b-c-a\right)^3+\left(c-a-b\right)^3\)

\(=\left(2\right)^3+\left(-2a\right)^3-6a\left[a+\left(b+c\right)\right]\left[a-\left(b+c\right)\right]+6a\left[-a+\left(b-c\right)\right]\left[-a-\left(b-c\right)\right]\)

\(=-6a\left\{a^2-\left(b+c\right)^2-\left[\left(-a\right)^2-\left(b-c\right)^2\right]\right\}\)

\(=-6a\left\{a^2-a^2+\left(b-c\right)^2-\left(b+c\right)^2\right\}\)

\(=-6a\left[b-c+b+c\right]\left[b-c-\left(b+c\right)\right]=-6a.2b.\left(-2c\right)\)

\(=24abc\)

ừ chie cần k vaod chữ đúng thôi

a,Đặt a+b-c=x, c+a-b=y, b+c-a=z

=>x+y+z=a+b-c+c+a-b+b+c-a=a+b+c

Ta có hằng đẳng thức:

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

=>(a+b+c)^3-(b+c-a)^3-(a+c-b)^3-(a+b-c)^3=(x+y+z)^3-x^3-y^3-z^3

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

=3(a+b-c+c+a-b)(c+a-b+b+c-a)(b+c-a+a+b-c)

=3.2a.2b.2c

=24abc

mình mới có tài khoản,vậy k cho bn chỉ cần k đúng thôi đk ^^?