IF x+y=1 then `x^3+y^3+3xy-1`=

Options

A. 0
B. 5
C. 2
D. 1

Answer

To solve the equation \( x^3 + y^3 + 3xy - 1 \) given that \( x + y = 1 \), we can follow these steps:

### Step 1: Use the identity for the sum of cubes
We know from algebra that:
\[ x^3 + y^3 = (x+y)(x^2 - xy + y^2) \]
Since we have \( x + y = 1 \), we can substitute this into the equation.

### Step 2: Substitute \( x + y \)
Substituting \( x + y = 1 \) into the identity gives:
\[ x^3 + y^3 = 1 \cdot (x^2 - xy + y^2) = x^2 - xy + y^2 \]

### Step 3: Express \( x^2 + y^2 \)
We can express \( x^2 + y^2 \) in terms of \( (x+y)^2 \):
\[ x^2 + y^2 = (x+y)^2 - 2xy = 1^2 - 2xy = 1 - 2xy \]
Thus, we can rewrite \( x^3 + y^3 \):
\[ x^3 + y^3 = (1 - 2xy) - xy = 1 - 3xy \]

### Step 4: Substitute back into the original expression
Now we substitute \( x^3 + y^3 \) back into the original expression:
\[ x^3 + y^3 + 3xy - 1 = (1 - 3xy) + 3xy - 1 \]

### Step 5: Simplify the expression
Now simplify the expression:
\[ (1 - 3xy + 3xy - 1) = 0 \]

### Conclusion
Thus, the value of \( x^3 + y^3 + 3xy - 1 \) is:
\[ \boxed{0} \]