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

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

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt :33

Thank you))))

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

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

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

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

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

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

Giờ là cách khác:(tại em làm khá kĩ nên nó dài thôi chứ em trình bày lại trong giấy nó ngắn ngủn à)

Đặt \(b^2-c^2=x;c^2-a^2=y\Rightarrow a^2-b^2=-\left(x+y\right)\)

Suy ra \(M=ax+by-c\left(x+y\right)\)

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

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

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

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

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

\(=\left(b-c\right)\left(a-c\right)\left(b-a\right)\) [muốn cho đẹp thì nhân (-1) . (-1) vào thì nó thành (a-b)(b-c)(c-a) ]

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

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

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

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

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

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

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

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

Mình đã làm bài này rồi.

Link: https://hoc24.vn/hoi-dap/question/824554.html

(x -y)³  - 1 - 3(x -y)(x - y - 1)

= (x -y)³  - 3(x -y)(x - y - 1) - 1

Đặt x - y = t, khi đó ta có:

    t³  -  3t. (t  - 1) - 1

=   t³  -  3t²  + 3t - 1

=  (t  - 1)³

Thay t = x - y vào (t - 1)³ , ta có:  ( x - y - 1)³

Vậy (x -y)³  - 1 - 3(x -y)(x - y - 1) =  ( x - y - 1)³

Câu a :

\(\left(x-5\right)^2+\left(x-5\right)\left(x+5\right)-\left(5-x\right)\left(2x+1\right)\)

\(=x^2-10x+25+x^2-25-10x-5+2x^2+x\)

\(=4x^2-19x-5\)

Câu b :

\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

\(=12x^2-9x-8x+6-2x+2+3x^2-3x-6x^2-6x+4x+4\)

\(=9x^2-24x+2\)