Find 2 consecutive integers, the smaller a perfect cube and larger a perfect square.
8 = 2^3
9 = 3^2
Thank you Ms.Sue!!!!
You're welcome, Marissa. :-)
To find two consecutive integers, where the smaller one is a perfect cube and the larger one is a perfect square, we can start by understanding what a perfect cube and a perfect square are.
A perfect cube is a number that can be expressed as the cube of an integer, while a perfect square is a number that can be expressed as the square of an integer.
Let's assume the smaller number as 'x'. Since 'x' is a perfect cube, we can express it as x = a^3, where 'a' is an integer.
Next, we need to find the larger number, which is a perfect square. Let's assume it as 'y'. Therefore, we can express it as y = b^2, where 'b' is an integer.
Now, since we are looking for two consecutive integers, we can say that y = x + 1.
Replacing y with its expression, we have:
b^2 = a^3 + 1
To find the consecutive integers, we can try different values of 'a' and see if there is any corresponding value of 'b' that satisfies the equation.
Let's try 'a' starting from 1 and increment it until we find a solution.
For a = 1, we have:
b^2 = 1^3 + 1
b^2 = 2
This equation doesn't have an integer solution for 'b'.
Trying with a = 2:
b^2 = 2^3 + 1
b^2 = 9
For b = 3, this equation is satisfied.
Therefore, if we take x as 8 (2^3) and y as 9 (3^2), we have found two consecutive integers where the smaller one is a perfect cube (8) and the larger one is a perfect square (9).