a ) Khai triển : \(\left(x+y\right)^5\) theo nhị thức Newton , ta có :
Đặt \(A=\left(x+y\right)^5-x^5-y^5\)
\(=5x^4y+10x^3y^2+10x^2y^3+5xy^4\)
\(=5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
Mà \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right);2x^2y+2xy^2=2xy\left(x+y\right)\)
Do đó : \(A=5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
b ) Đặt \(B=a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+cc^2\left(c-a-b\right)\)
\(=3abc+a^3+b^3+c^3-a^2b-b^2a-a^2c-b^2c-c^2a-c^2b\)
\(=a^2\left(a-b\right)+b^2\left(b-a\right)+c\left(2ab-a^2-b^2+c\left(c^2-bc-ac+ab\right)\right)\)
\(=\left(a-b\right)\left(a^2-b^2\right)-c\left(a-b\right)^2+c\left(c-a\right)\left(c-b\right)\)
\(=\left(a-b\right)^2\left(a+b+c\right)+c\left(b-c\right)\left(a-c\right)\)
\(A=B-c\left(b-c\right)\left(a-c\right)=\left(a+b\right)^2\left(a+b-c\right)\).
