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

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

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

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

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

a) \(A=x^3+y^3+3xy\)

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

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

\(=\left(x+y\right)^3\) \(=1\)

b) \(B=x^3-y^3-3xy\)

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

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

\(=\left(x-y\right)^3\) \(=1\)

\(a,x+y=1\Leftrightarrow\left(x+y\right)^3=1\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\\ \Leftrightarrow x^3+y^3+3xy\cdot1=1\Leftrightarrow x^3+y^3+3xy=1\)

\(b,x^3-y^3-3xy\\ =x^3-3x^2y+3xy^2-y^3-3xy+3x^2y-3xy^2\\ =\left(x-y\right)^3-3xy\left(x-y-1\right)\\ =1^3-3xy\left(1-1\right)=1-0=1\)

\(c,x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\\ =x^2-xy+y^2+3xy-6x^2y^2+6x^2y^2\\ =x^2+2xy+y^2=\left(x+y\right)^2=1\)

`#3107.101107`

`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`

Ta có:

`x - y - 1 = 0`

`=> x - y = 1`

`D = x^3 - y^3 - 3xy`

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

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

`= x^2+ xy + y^2 - 3xy`

`= x^2 - 2xy + y^2`

`= x^2 - 2*x*y + y^2`

`= (x - y)^2`

`= 1^2 = 1`

Vậy, với `x - y = 1` thì `D = 1`

________

`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`

`x + y = 5`

`=> (x + y)^2 = 25`

`=> x^2 + 2xy + y^2 = 25`

`=> 2xy = 25 - (x^2 + y^2)`

`=> 2xy = 25 - 17`

`=> 2xy = 8`

`=> xy = 4`

Ta có:

`E = x^3 + y^3`

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

`= 5 * [ (x^2 + y^2) - xy]`

`= 5 * (17 - 4)`

`= 5 * 13`

`= 65`

Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`

________

`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`

Ta có:

`x - y = 4`

`=> (x - y)^2 = 16`

`=> x^2 - 2xy + y^2 = 16`

`=> (x^2 + y^2) - 2xy = 16`

`=> 2xy = (x^2 + y^2) - 16`

`=> 2xy = 26 - 16`

`=> 2xy = 10`

`=> xy = 5`

Ta có:

`F = x^3 - y^3`

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

`= 4 * [ (x^2 + y^2) + xy]`

`= 4 * (26 + 5)`

`= 4*31`

`= 124`

Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`

a) A = -1;                        b) B = ( x   +   y ) 3  =1.

Ta có:

\(x+y=1\Rightarrow3xy=3xy\left(x+y\right)\)

\(x^3+3xy+y^3\)

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

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

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

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

Q =  x - y 3  +  y + x 3  +  y - x 3  – 3xy(x + y)

      = x 3  – 3 x 2 y + 3x y 2  – y 3  +  y 3  + 3 y 2 .x + 3y x 2  +  x 3  + y 3  – 3 y 2 .x +3y x 2  –  x 3  – 3 x 2 y – 3x y 2

      =  x 3  – 3 x 2 y + 3x y 2  –  y 3  +  y 3  + 3.x y 2  + 3 x 2 .y +  x 3  +  y 3  – 3x. y 2 + 3 x 2 .y –  x 3  – 3 x 2 y – 3x y 2

       = (  x 3  +  x 3  –  x 3 )+ ( - 3 x 2 y + 3 x 2 y+ 3 x 2 y – 3 x 2 y)+ (3x y 2  + 3x y 2  - 3x y 2 - 3x y 2 ) + (- y 3 +  y 3 +  y 3  )

       =  x 3  + 0 x 2 y + 0.x y 2  +  y 3

       =  x 3 + y 3