Given that the sum of 2 numbers is 1 and their product is 1, let's find the sum of their cubes.
Let x and y represent the numbers.
Thus, we have:
x + y = 1..........................equation 1
x y = 1...........................equation 2
Here, we are to find x³ + y³ .
Thus, we can write this as:
![x^3+y^3=(x+y)(x^2+y^2-xy)](https://img.qammunity.org/2023/formulas/mathematics/college/wuzx99qnj3qhcgzipoqufe73el8xgn1dyz.png)
Manipulating this, we can write:
![x^3+y^3=(x+y)((x+y)^2-3xy))](https://img.qammunity.org/2023/formulas/mathematics/college/60dq7rg7ahsznckbnz9debde9lff5duy0c.png)
Substitute the values for x+y and xy into the equation above, we have:
![\begin{gathered} x^3+y^3=(1)((1)^2-3(1)) \\ \\ x^3+y^3=(1)(1-3) \\ \\ x^3+y^3=-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/51vbfbha7n6o3dyar899dx229f2vaos41t.png)
Therefore, the sum of their cubes is -2.
ANSWER:
-2