Giúp tuii với huhuu

Ta có

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

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

Áp dụng vào tổng ta có

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

Đặt

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

=> M=0

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

\(\Rightarrow N=3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

Để ý : M+N=B

=> \(B=0+3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

=> \(B=3\left(c-a\right)\left(a-b\right)\left(b-c\right)\)

❤

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Bạn tham khảo bài trong hình nhé. Nếu không thấy hình thì vào câu hỏi tương tự để xem.

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

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

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

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

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

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

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

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(5c-a-4b\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(b+c-2a\right)\left(a+4b-5c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(a+4b-5c-b-c+2a\right)\)

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

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

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

Đặt: \(a+b-2c=x;b+c-2a=y;c+a-2b=z\)

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

Ta thấy: \(x+y+z=a+b-2c+b+c-2a+c+a-2b=0\)

\(x+y=a+b-2c+b+c-2a=2b-a-c\)

\(y+z=b+c-2a+c+a-2b=2c-a-b\)

\(z+x=c+a-2b+a+b-2c=2a-b-c\)

Thay vào B \(\Rightarrow B=0-3\left(2b-a-c\right)\left(2c-a-b\right)\left(2a-b-c\right)\)

Vậy \(B=-3\left(2b-a-a\right)\left(2c-a-b\right)\left(2a-b-c\right).\)

Đặt \(a+b-2c=x,b+c-2a=y,c+a-2b=z\)

\(\Rightarrow x+y+z=0\)

Chắc bạn biết: \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

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

Chúc bạn học tốt.

Đặt:  \(a+b-2c=x;\)   \(b+c-2a=y;\)\(c+a-2b=z\)

=>   \(x+y+z=0\)

=>  \(x^3+y^3+z^3=3xyz\)

Thay trở lại ta được:

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

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