x+y=1 => (x+y)^3=1 <=> x^3+3x^2y+3xy^2+y^3=1

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

<=> x^3+y^3+3xy=1 Do x+y=1

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

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

Bài này là trên vio mk cx gặp r

x³+y³+3xy

= (x+y) ( x²-xy+y2) + 3xy

= 1 .( x²-xy+y²)+ 3xy

= x²-xy+y²+3xy

= x²+2xy+y²

=( x+y )²

= 1²

=1

lik e nha

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\(xy\)

ta có : x³ + y³ + 3xy = (x+y)(x² -xy +y²) +3xy =x²-xy+y²+3xy= (x+y)²=1

Ta co.                  

X^3+y^3=1 

=1+3xy

=4xy

bn Toàn đúng, đây là bài thi violympic v3, mk thi rùi

Ta có

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

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

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

\(\Rightarrow K=1^3-3xy+3xy=1\)

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)  \(x-y=1\)

=>  \(\left(x-y\right)^3=1\)

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

<=>  \(x^3-y^3-3xy=1\)