a ) Ta có :
\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(=\left(a^2+b^2+2ab\right)+\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
Vậy ...
b) Ta có :
\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(\Rightarrow\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2-a^3-b^3+c^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2-\left(a+b\right)\left(a^2+b^2-ab\right)\)
\(=\left(a+b\right)\left[\left(a+b\right)^2+3\left(a+b\right)c+3c^2-a^2-b^2+ab\right]\)
\(=\left(a+b\right)\left[a^2+b^2+2ab+3ac+3bc+3c^2-a^2-b^2+ab\right]\)
\(=\left(a+b\right)\left[3ab+3ac+3bc+3c^2\right]\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
Vậy ...
