Ta có: A=(a+b+c)3+(a−b−c)3−6a(b+c)2 

= a^3  + b^3 + c^3 + 3ab + 3ac + 3bc + a^3 + b^3 + c^3 - 3ab - 3ac + 3bc - 6a(b^2+2bc + c^2)

= a^3  + b^3 + c^3 + 3ab + 3ac + 3bc + a^3 + b^3 + c^3 - 3ab - 3ac + 3bc - 6ab^2 + 12abc+6ac^2

=2a^3 + 2b^3 + 2c^3 + 6a^2 + 12abc

Cậu dùng hằng đẳng thức nâng cao là ra. Nhớ tick mình nha,

Đặt b+c=x ta được:

A=(a+x)³+(a-x)³-6a.x²

=a³+3a²x+3ax²+x³+a³-3a²x+3ax²-x³-6ax²

=2a³

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

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

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

\(=2a^3\)

kết quả là A=\(2a^3\)

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

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

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

bài trên nhé

=a^3+b^3+c^3+a^3-b^3-c^3-(6ab+6ac)^2

=2a^3-6ab^2-6ac^2