If x + y = 1, then find the value of $$x^3+y^3+3xy.$$​​

Correct option is C

Given:

We are given the equation $x+y=1$ and are asked to find the value of $$x^3 + y^3 + 3xy$$ 

Formula Used:

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

$$(x + y)^2 = x^2 + 2xy + y^2$$​​

Solution:

Substitute x + y =1 into this identity:

​$$x^3 + y^3 = (1)(x^2 - xy + y^2) = x^2 - xy + y^2$$ 

We know,

​$$(x + y)^2 = x^2 + 2xy + y^2$$ 

Since $x+y=1$, we have:

$$1^2 = x^2 + 2xy + y^2$$ 

$$1 = x^2 + 2xy + y^2$$ 

​$$x^2 + y^2 = 1 - 2xy$$ 

Thus, we have: 

​$$x^2 - xy + y^2 = (x^2 + y^2) - xy = (1 - 2xy) - xy = 1 - 3xy$$ 

Now substituting  $$x^3 + y^3 = 1 - 3xy$$​ into the original expression $$x^3 + y^3 + 3xy$$ 

$$x^3 + y^3 + 3xy = (1 - 3xy) + 3xy$$ 

​$$x^3 + y^3 + 3xy = 1$$ 

The value of ($$x^3 + y^3 + 3xy$$​) is: 1.

​