bn muốn hỏi gì vậy ????

nhưng sao ra đc nhận tủ chung chứ

dòng thứ 3 ý

1) \(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)

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

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

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

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

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

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

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

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

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

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

Phân phối ra rồi rút gọn thôi: \(24abc\)

bn có toán nâng cao và phát triển ko trong đó có đấy

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

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

= ( a + b )³ + 3.( a + b )².c +  3.( a + b ).c² + c³ + ( a - b )³ - 3.( a - b )².c + 3.( a - b ).c² - c³ + ( - c³ ) + 3.( a - b )².c - 3.( a - b ).c² -(a- b)³

+ c³ + 3.( a + b )².c - 3.( a + b ).c² - ( a + b )³

= 6.( a + b )² .c 

Đặt A là tên biểu thức; \(a+b-c=x;b+c-a=y;c+a-b=z\)

Khi đó \(x+y+z=a+b-c+b+c-a+c+a-b=a+b+c\)

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

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

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

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

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

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

\(=3.2b.2c.2a=24abc\)

a)4y(x-1)-(x+1)=4y(x-1)-(X-1)

   =(x-1) (4y-1)

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

=x-3 (x-3)^2

HOK TỐT K NHA

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

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

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

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

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

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

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

a) a³ (b-c) + b³ (c-a) +c³ (a-b)

<=> a³b – a³c +b³c – b³a + c³a – c³b

<=>  b(a³ – c³) +c(a³ – b³) + a(b³ - c³)

(Tự áp dụng hằng đẳng thức)

b)