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

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

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

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

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

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

                                                                                                                                 = 1^2 = 1 

Ta có : x³ - y³ = ( x - y )³ + 3xy(x - y) = 1 +3xy

=> x³ - y³ - 3xy = 1 + 3xy - 3xy = 1

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

\(A=x^3-y^3-3xy\)

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

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

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

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

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

=> \(A=x^3-y^3-3xy=\left(x-y\right)^3=1^3=1\)

Linh_Men

a)\(M=\text{[}x^3-3xy\left(x-y\right)-y^3\text{]}-\left(x^2-2xy+y^2\right)\)

\(M=\left(x-y\right)^3-\left(x-y\right)^2\)

\(\Rightarrow M=7^3-7^2\)

\(M=294\)

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=...

C1:  \(B=x^3+3xy+y^3\)

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

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

Thay \(x+y=1\)ta được:

\(B=1^3-3xy\left(1-1\right)=1\)

C2: \(x+y=1\)\(\Rightarrow\)\(x=1-y\)

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

\(=1-3y+3y^2-y^3+3y-3y^2+y^3=1\)

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

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

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

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

Mà x+y = 1 

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

Vậy \(N=x^3+y^3+3xy=1\)

Câu b làm tương tự bạn nhé !!

a,\(\left(x^3+y^3\right)=x^3+y^3+3x^2y+3xy^2\)

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

\(x+y=1\)

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

Vậy \(N=x^3+y^3+3xy=1\)

Bạn tự làm tiếp nha

BÀI 1:
   \(N=x^3+y^3+3xy=x^3+y^3+3xy\cdot1\)
 Thay (x + y) = 1 vào N, ta đc: 
\(N=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1\)

BÀI 2: (Bạn làm tương tự)
 \(M=x^3-y^3-3xy=x^3-y^3-3xy\left(x-y\right)=\left(x-y\right)^3=1\)