Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

Ta có: x^3 -3xy(x-y) -y^3 -x^2 + 2xy-y^2

= x^3 -y^3 - 3xy(x-y) -( x^2 -2xy+y^2)

= (x-y)(x^2+xy +y^2) - 3xy(x-y) -(x-y)^2

= (x-y)(x^2+xy+y^2 -3xy-x+y)

=11( x^2 -2xy+y^2 -x+y)

= 11[ (x-y)^2 -(x-y)]

= 11[ 11^2 -11]

= 11^3 -11^2=...

ukm, theo giả thiết , ta có y/5=z/5 => y=z sau đó bn tự hỉu nhé ^_^" 

giup minh vo nao thanhs

a/ \(M=x^4-xy^3+x^3y-y^4-1\)

\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)

Mà \(x+y=0\)

\(\Leftrightarrow M=x^3.0-y^3.0-1\)

\(\Leftrightarrow M=-1\)

Vậy ...

Q=2