Question

Show that (x-y)^3=x^3-y^3-3xy(x-y)​

Answer 1

Answer:Given that  x + y = 1, square both sides of the equation to get:

 

x^2 + 2xy + y^2 = 1.  

 

Now, multiply both sides of this equation by the quantity (x + y) to get the following:

 

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

 

Now, grouping the cubed terms on the left side together, we have the following:

 

x^3 + y^3 + 3x^2y +3xy^2 = x + y.  Now factoring on the left side, we have:

 

x^3 + y^3 + 3xy(x + y) = x + y.  Now, substituting by using the given condition that x + y =1,  we have:

 

x^3 + y^3 + 3xy(1) = 1,  and performing the simple multiplication by 1, we finally get:

 

x^3 + y^3 + 3xy = 1

Step-by-step explanation: