x³+y³+z³-3xyz

= (x+y)³-3xy(x+y)+z³-3xyz

= [(x+y)³+z³]-[3xy(x+y)+3xyz]

=(x+y+z)[(x+y)²-(x+y).z+z²]-3xy(x+y+z)

=(x+y+z)(x²+y²+z²+2xy-xz-yz) -3xy(x+y+z)

= (x+y+z)(x²+y²+z²-xy-xz-yz)

\(B=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-xz\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

\(+3xyz\)hay\(-3xyz\)vậy bạn

Goi da thuc tren la A

Thay a=b -> A= 0 -> A chua nghiem la a-b

Tuong tu b=c-> A = 0 - > A chua nghiem la b -c

Tuong tu c =a - > A = 0 -> A chua nghiem la c-a

=> A = k(a - b)(b - c)(c - a)

Vì A có bậc 3 mà (a - b)(b - c)(c - a) cũng có bậc 3 -> k là 1 số 

Thay a = 3, b= 2, c= 1

=> A= -6=k.1.1..-2

=> k = 3

=> A = 3(a - b)(b - c)(c - a)

Đây gọi là phương pháp giá trị riêng bạn nha!

x^5 + x + 1

= x^5 - x^2 + (x^2 + x + 1)

= x^2(x^3 - 1) + ( x^2 + x + 1)

= x^2( x - 1)(x^2 + x + 1) + ( x^2 + x + 1)

= (x^3 - x^2 + 1)(x^ 2 + x + 1)