a. \(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3-\left(c+a-b\right)^3\)
Đặt \(a+b-c=x\) , \(b+c=y,c+a-b=z\) thì:
\(x+y+z=a+b-c+b+c-a+c+a-b=a+b+c\)
Áp dụng hằng đẳng thức ta có:
\(\left(x+y+z\right)^3-x^3-y^3-z^3=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Ta có: \(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3\left(c+a-b\right)^3\)
\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\)
\(=3.2b.2c.2a=24abc\)
b. \(abc-\left(ab+bc+ca\right)\left(a+b+c\right)-1\)
\(=abc-bc-ab+b-ac+c+a-1\)
\(=bc\left(a-1\right)-b\left(a-1\right)-c\left(a-1\right)+\left(a-1\right)\)
\(=\left(a-1\right)\left(bc-b-c+1\right)=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
a, Đặt \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Rightarrow x+y+z=a+b+c\)
\(\Rightarrow A=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(y+z\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+xy+xz+x^2-y^2+yz-z^2\right)\)
\(=\left(y+z\right)\left(3x^2+3xy+3yz+3xz\right)\)
\(=3\left(y+z\right)\left(x^2+xy+yz+xz\right)\)
\(=3\left(y+z\right)\left[x\left(x+z\right)+y\left(x+z\right)\right]\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
\(=24abc\)
Vậy A = 24abc
