Question

phân tích đa thức thành nhân tử

\(A\left(a;b;c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

Answer 1

A = a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc

   = (a^2b+ab^2+abc)+(b^2c+bc^2+abc)+(c^2a+ca^2+abc)

   = (a+b+c).(ab+bc+ca)

k mk nha

Answer 2

( a + b ) ( b + c ) ( c + a ) + abc 

= ( ab + ac + b^2 + bc ) ( c + a ) + abc

= ( c + a ) ( ab + ac + b^2 + bc ) + abc

= abc + ac^2 + b^2c  + bc^2 + a^2b + a^2c + ab^2 + abc + abc 

= 3abc + a . c^2 + b^2 . c + b . c^2 + a^2 . b + a^2 . c + a . b^2 

= 3abc + c^2( a + b ) + b^2( c + a ) + a^2 ( b + c )