Ta có : Thêm \(-3xyz\) vào 2 vế , ta có :

\(VT=x^3+y^3+z^3-3xyz\)

       \(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\left(1\right)\)

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

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

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

Từ ( 1 ) và ( 2 ) \(\Rightarrow x^3+y^3+x^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3\)

\(\Rightarrowđpcm\)

Cho sủa đề nha : \(x^3+y^3+x^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3\)

Ta có : x + y = -1
=> ( x + y )² = 1
=> - ( x + y )² = -1
=> - ( x² + 2xy + y² ) = -1
=> -x² - 2xy - y² = -1
=> - x² + xy - y² - 3xy = -1
=> -( x² - xy + y² ) - 3xy = -1
=> -1 . ( x² - xy + y² ) - 3xy = -1
Thay -1 = x + y vào biểu thức ta có :

( x + y ) . ( x² - xy + y² ) - 3xy = -1
=> x³ + y³ - 3xy = -1 ( ĐPCM )

Câu a : Ta có :

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

Câu b : Ta có :

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

a) Vì x + y = 1 => ( x + y )³ = 1

=> x³ + 3x²y + 3xy² + y³ = 1

=> x³ + y³ + 3xy ( x + y ) = 1

=> x³ + y³ +3xy = 1 (do x+y=1)

b) x-y=1 => (x-y)³=1

=> x³ - 3x²y + 3xy² -y³ = 1

=> x³ -y³ - 3xy (x - y) = 1 

=> x³ - y³ -3xy =1 (do x-y=1) 

x + y = 1

=> (x + y)³ = 1

<=> x³ + y³ + 3x²y + 3xy² = 1

<=> x³ + y³ + 3xy (x+y) = 1

<=> x³ + y³ + 3xy = 1

Vậy ... = 1

x - y = 1

=> (x - y)³ = 1

<=> x³ - y³ - 3x²y + 3xy² = 1

<=> x³ - y³ - 3xy (x - y) = 1

<=> x³ - y³ - 3xy = 1

Vậy ... = 1

1) \(A=x^3+y^3+3xy\)

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

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

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

Vậy A = 1.

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

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

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

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

\(B=\left(x-y\right)^2=1^2=1\)

Vậy B = 1.

Ta có :

\(VT=\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(y+x\right)}{\left(x-y\right)^3}\)

\(VT=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}=\frac{-x-y}{\left(x-y\right)^2}=VP\)

Vậy .......................

a) Ta có:

\(x+y=1\)

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

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

\(\Rightarrow x^3+y^3+3xy=1\)

\(\Rightarrow P=1\)

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

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

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

Thay x + y = 1 vào Q

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

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

\(Q=1\)