This problem describes a sequence of squaring operations that results in the final number 6561.

We can break the problem into steps:

Let the original number be $x$.

1. The number is squared: $x^2$.
2. The result is squared again: $(x^2)^2 = x^4$.
3. The result is squared one more time: $(x^4)^2 = x^8$.

The final result is 6561, so we have the equation: $x^8 = 6561$

To find $x$, we take the eighth root of 6561: $x = \sqrt[8]{6561}$

Now, let's calculate the eighth root of 6561.

I'll compute this now.The original number $x$ is $3$.

Now, to answer each part of the question:

a) The original number is $3$.

b) The next two numbers are:

• First, squaring the original number: $3^2 = 9$
• Then, squaring that result: $9^2 = 81$

c) Among these numbers (3, 9, and 81), both 9 and 81 are perfect squares.

Would you like further details or have any questions? Here are some related questions:

1. How do we find the square and eighth roots of larger numbers?
2. Can squaring operations be represented visually with functions?
3. What is the difference between perfect squares and other numbers?
4. How do powers and exponents interact in sequences like this?
5. What would happen if you continued squaring numbers in this sequence?

Tip: Always verify the final result by working backward to ensure your solution is correct.