b; 1³ = (\(x-y\))³ = \(x^3\) - 3\(x^2\).y + 3\(xy^2\) - y³ = \(x^3\) - y³ - 3\(xy\)(\(x-y\)) 

    1 = \(x^3\) - y³ - 3\(xy\)

a)  \(x+y=1\)

=>   \(\left(x+y\right)^3=1\)

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

<=>  \(x^3+y^3+3xy=1\)

b)  \(x-y=1\)

=>  \(\left(x-y\right)^3=1\)

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

<=>  \(x^3-y^3-3xy=1\)

1³ = (\(x+y\))³ = \(x^3\) + 3\(x^2\)y + 3\(xy^2\) + y³ = \(x^3\)+y³+3\(xy\)(\(x+y\))

1 = \(x^3\)+y³+3\(xy\)

1³ = (\(x-y\))³ = \(x^3\) - 3\(x^2\)y + 3\(xy\) - y³ = \(x^3\) - y³ - 3\(xy\)(\(x-y\))

1 = \(x^3\) - y³ - 3\(xy\)

   \(A=x^3+3xy+y^3\)

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

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

       \(=x^2-xy+y^2+3xy\)

       \(=x^2+2xy+y^2\)

       \(=\left(x+y\right)^2\)

       \(=1\)

x³+y³=x³+3x²y+3xy²+y²+3xy-3x²y-3xy²

=(x+y)³+3xy.(1-x-y)

=(x+y)³+3xy.[1-(x+y)]

=1³+3xy.(1-1)

=1

1³ - 3xy . (1-1) = 1 

>_< chúc bn học tốt

1, \(A=x^3+y^3+3xy\)

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

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

Thay x +1 = 1 ta có 

\(1^3+3xy-3xy.1=1+3xy-3xy=1\)

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 \(x-y=1\))

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

\(=\left(x-y\right)^3\) \(=1\)