\(A=\left(a-1\right)^3+\left(b-2\right)^3+\left(c-3\right)^3\)
\(\Rightarrow A-3.\left(a-1\right)\left(b-2\right)\left(c-3\right)=\)\(\left(a-1\right)^3+\left(b-2\right)^3+\left(c-3\right)^3-3\left(a-1\right)\left(b-2\right)\left(c-3\right)\)
Đặt \(\left\{{}\begin{matrix}a-1=x\\b-2=y\\c-3=z\end{matrix}\right.\)
Thay \(\left(a-1\right)\left(b-2\right)\left(c-3\right)=100\)vào \(\Leftrightarrow A-300=x^3+y^3+z^3-3xyz\)
Phân tích \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
(Link để bạn hiểu cái trên: a^3+b^3+c^3 - Tìm với Google)
\(A-300=\left(a-1+b-2+c-3\right)\).\(\left[\left(a-1\right)^2+\left(b-2\right)^2+\left(c-3\right)^2-\left(a-1\right)\left(b-2\right)-\left(b-2\right)\left(c-3\right)-\left(a-1\right)\left(c-3\right)\right]\)
\(\Rightarrow A-300=0\Rightarrow A=300\)
Vậy...
@Nguyễn Thị Ngọc Thơ
\(a+b+c=6\Rightarrow\left(a-1\right)+\left(b-2\right)+\left(c-3\right)=0\)
Ta có mệnh đề: \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
\(\Rightarrow\left(a-1\right)^3+\left(b-2\right)^3+\left(c-3\right)^3=3\left(a-1\right)\left(b-2\right)\left(c-3\right)=300\)