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

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

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

\(=\left[a^3+b^3+3ab.\left(a+b\right)+c^3+3c.\left(a+b\right)\right]-a^3-b^3-c^3\)

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

Áp dụng :

Đặt \(\left\{{}\begin{matrix}a+b-c=x\\a-b+c=y\\-a+b+c=z\end{matrix}\right.\) \(\Rightarrow x+y=z=a+b+c\)

Khi đó biểu thức trở thành :

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

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

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

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

\(=\left[\left(a+b\right)^3+3.\left(a+b\right)^2.c+3.\left(a+b\right).c^2+c^3\right]-\left[\left(b+c\right)^3-3.\left(b+c\right)^2.a+3.\left(b+c\right).a^2-a^3\right]-\left[\left(a+c\right)^3-3.\left(a+c\right)^2.b+3.\left(a+c\right).b^2-b^3\right]-\left[\left(a+b\right)^3-3.\left(a+b\right)^2.c+3.\left(a+b\right).c^2-c^3\right]\)\(=\left[\left(a^3+3a^2b+3ab^2+b^3\right)+3\left(a^2+2ab+b^2\right).c+3c^2a+3c^2b+c^3\right]-\left[\left(b^3+3b^2c+3bc^2+c^3\right)-3.\left(b^2+2bc+c^2\right).a+3a^2b+3a^2c-a^3\right]-\left[\left(a^3+3a^2c+3ac^2+c^3\right)-3\left(a^2+2ab+b^2\right).c+3c^2a+3c^2b-c^3\right]\)\(=\left(a^3+3a^2b+3ab^2+b^3+3ca^2+6abc+3b^2c+3c^2a+3c^2b+c^3\right)-\left(b^3+3b^2c+3bc^2+c^3-3ab^2-6abc-3ac^2+3a^2b+3a^2c-a^3\right)-\left(a^3+3a^2c+3ac^2+c^3-3a^2c-6abc-3b^2c+3c^2a+3c^2b-c^3\right)\)\(=a^3+3a^2b+3ab^2+b^3+3ca^2+6abc+3b^2c+3c^2a+3c^2b+c^3-b^3-3b^2c-3bc^2-c^3+3ab^2+6abc+3ac^2-3a^2b-3a^2c+a^3-a^3-3a^2c-3ac^2-c^3+3a^2c+6abc+3b^2c-3c^2a-3c^2b+c^3\)\(=3ab^2+6abc+3ab^2+6abc+a^3+6abc+3b^2c-3c^2b\)

\(=6ab^2+18abc+a^3+3b^2c-3bc^2\)

P/s: Ko chắc! Đây là kết quả của hơn 50 phút !

a. Câu hỏi của Nhàn Nguyễn - Toán lớp 8 - Học toán với OnlineMath

a) Đặt a+b-c=x , b+c-a=y, c+a-b=z

⇒(a+b+c)³−x³−y³−z³

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

⇒(x+y+z)³−x³−y³−z³

=[(x+y)+z³]−x³−y³−z³

=(x+y)³+z³+3^z(x+y)(x+y+z)−x³−y³−z³

=x³+y³+3xy(x+y)+z³+3z(x+y)(x+y+z)−x³−y³−z³

=3(x+y)(xy+xz+yz+z²)

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

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

Áp dụng hằng đẳng thức trên ta có

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

= 3.2b.2c.2a

= 24abc

K MIK NHA BẠN ^^

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

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

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

\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(\text{đ}pcm\right)\)