a)\(a^3+b^3+c^3-3abc\)

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

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

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

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

cho tam giac abc goi d,e lan luot la trung diem ab,ac.g la giao diem be,cd.goi m doi xung vs c qua d,n doi xung  vs b qua e vs i,k lan luot doi xung vs g qua d,e

a)cm n,m doi xung vs nhau qua a

b)tinh de,biet bc=3cm

Ta có:\(x^3-7x-6=\left(x^3-3x^2\right)+\left(3x^2-9x\right)+\left(2x-6\right)\)

\(=\left(x-3\right)\left(x^2+3x+2\right)=\left(x-3\right)\left(x^2+2x+x+2\right)\)

\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)

=x³-x-6x-6

=(x³-x)-(6x-6)

=x(x²-1)-6(x-1)

=x(x-1)(x+1)-6(x-1)

=(x-1)(x²+1-6)

                                         Bài làm :

                          = x² - 2x - 4x + 8

                          = x (x - 2) - 4(x -2)

                          = (x - 4)(x -2)

                     = x² - 6x + 9 - 1

                     = ( x - 3)² - 1

                     =( x -3 - 1)( x- 3 + 1)

                     = (x - 4)(x -2)

                       = x² - 16 - 6x + 24

                       =( x - 4)(x + 4 ) - 6 (x - 4)

                       =(x - 4)(x + 4 - 6)

                       = (x - 4)(x -2) 

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

mình cũng được tròn 3 cách 

c1 \(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

c2 \(x^2-6x+8=\left(x^2-6x+9\right)-1=\left(x-3\right)^2-1=\left(x-4\right)\left(x-2\right)\)

c3 Gỉa sử \(x^2-6x+8=\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)

Cân bằng hệ số ta được \(\hept{\begin{cases}a+b=-6\\ab=8\end{cases}< =>\orbr{\begin{cases}a=-4\\b=-2\end{cases}or\orbr{\begin{cases}a=-2\\b=-4\end{cases}}}}\)

Vậy ta có : \(\left(x+a\right)\left(x+b\right)=\left(x-2\right)\left(x-4\right)\)

Cách 1:

\(x^2-6x+8\)

\(=\left(x^2-4x\right)-\left(2x+8\right)\)

\(=x\left(x-4\right)-2\left(x-4\right)\)

\(=\left(x-2\right)\left(x-4\right)\)

Cách 2:

\(x^2-6x+8\)

\(=\left(x^2-2x\right)-\left(4x-8\right)\)

\(=x\left(x-2\right)-4\left(x-2\right)\)

\(=\left(x-2\right)\left(x-4\right)\)

Cách 3:

\(x^2-6x+8\)

\(=\left(x^2-6x+9\right)-1\)

\(=\left(x-3\right)^2-1\)

\(=\left(x-3+1\right)\left(x-3-1\right)\)

\(=\left(x-2\right)\left(x-4\right)\)

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2=0\)

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

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

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\) (luôn đúng vì \(a+b+c=0\))

Vậy \(a^3+b^3+c^3=3abc\)