Ta có : \(\left(a+b+c\right)^3-4\left(a^3+b^3+c^3\right)-12abc\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)-4\left(a^3+b^3+c^3\right)-12abc\)

\(=-3\left(a^3+b^3+c^3\right)+3\left(a+b\right)\left(a+c\right)\left(b+c\right)-12abc\)

\(=-3\left[a^3+b^3+c^3-\left(a+b\right)\left(a+c\right)\left(b+c\right)+4abc\right]\)

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

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

Giải quyết bằng toán này bằng cách đặt ẩn phụ. 

                                        \(--------------\)

Đặt  \(a+b=m\)   \(;\) \(a-b=n\)  thì  \(4ab=\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)=\left(a+b\right)^2-\left(a-b\right)^2\) , tức là  \(4ab=m^2-n^2\) và \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left[\left(a^2-2ab+b^2\right)+ab\right]=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\) ,

tức là  \(a^3+b^3=m\left(n^2+\frac{m^2-n^2}{4}\right)\)

Ta có:

\(A=\left(a+b+c\right)^3-4\left(a^3+b^3+c^3\right)-12abc\)

\(=\left(m+c\right)^3-4\left[m\left(n^2+\frac{m^2-n^2}{4}\right)+c^3\right]-3c\left(m^2-n^2\right)\)

\(=m^3+3m^2c+3mc^2+c^3-4mn^2-m^3+mn^2-4c^3-3m^2c+3n^2c\)

\(=3mc^2-3c^3-3mn^2+3n^2c\)

\(=3\left(mc^2-c^3-mn^2+n^2c\right)\)

\(=3\left[c^2\left(m-c\right)-n^2\left(m-c\right)\right]\)

\(=3\left(m-c\right)\left(c^2-n^2\right)=3\left(m-c\right)\left(c-n\right)\left(c+n\right)\)

Do đó,   \(A=3\left(a+b-c\right)\left(c-a+b\right)\left(c+a-b\right)\)

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

bai toan nay kho qua

(a+b+c)^3 thì viết được thành [(a+b)+c)]^3 rồi AD hằng đẳng thức để tính. Còn với (a^3+b^3+c^3) ta viết được (a+b)^3 -3a^2b -3ab^2 + c^3=(a+b)^3 -3ab(a+b)+c^3 ...thay vào rồi đổi biến

 k bt nhoak

Đặt \(a+b=m;a-b=n\)

Ta có:\(\Rightarrow\hept{\begin{cases}\left(a+b\right)^2=m^2\\\left(a-b\right)^2=n^2\end{cases}}\Rightarrow\hept{\begin{cases}a^2+2ab+b^2=m^2\\a^2-2ab+b^2=n^2\end{cases}}\Rightarrow\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)=m^2-n^2\)

\(\Rightarrow4ab=m^2-n^2\)

Mặt khác :\(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=m\left(n^2+\frac{m^2+n^2}{4}\right)\)

Ta lại có:\(A=\left(a+b+c\right)^3-4\left(a^3+b^3+c^3\right)-12abc\)

\(=\left(m+c\right)^3-4\left[m\left(n^2+\frac{m^2-n^2}{4}\right)+c^3\right]-12abc\)

\(=m^3+3m^2c+3c^2m+c^3-4\left(mn^2+\frac{m^2-n^2}{4}+c^3\right)-12abc\)

\(=m^3+3m^2c+3c^2m+c^3-4\left(\frac{4mn^2+m^3-mn^2}{4}+c^3\right)-3c\left(m^2-n^2\right)\)

\(=m^3+3m^2c+3c^2m+c^3-4\cdot\frac{m^3+3mn^2}{4}-4c^3-3cm^2+3cn^2\)

\(=m^3+3cm^2+3c^2m+c^3-m^3-3mn^2-4c^3-3cm^2+3cn^2\)

\(=\left(m^3-m^3\right)+\left(3cm^2-3cm^2\right)+3c^2m+\left(c^3-4c^3\right)+3cn^2-3mn^2\)

\(=3c^2m-3c^3+3cn^2-3mn^2\)

\(=3\left(c^2m-c^3+cn^2-mn^2\right)\)

\(=3\left[c^2\left(m-c\right)+n^2\left(c-m\right)\right]\)

\(=3\left(c^2-n^2\right)\left(m-c\right)\)

\(=3\left(c-n\right)\left(c+n\right)\left(m-c\right)\)

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

P/S:Bài giải dài.có j sai thông cảm cho e nha!

a) (x + 1)(x + 2)(x + 3)(x + 4) - 24

= [(x + 1)(x + 4)].[(x + 2)(x + 3)] - 24

= (x² + 5x + 4)(x² + 5x + 6) - 24 

= (x² + 5x + 5 - 1)(x² + 5x + 5 + 1) - 24

= (x² + 5x + 5)² - 1 - 24 = (x² + 5x + 5)² - 25 

= (x² + 5x)(x² + 5x + 10) 

 = x(x + 5)(x² + 5x + 10)

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

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

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

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

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

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