A Number Equal to Its Cube but Not to Its Square

In mathematics, there exists a fascinating property where a number can satisfy the condition of being equal to its cube but not to its square. Let's explore this intriguing concept through a detailed analysis.

Introduction to the Problem

Consider a number x that satisfies these conditions:

x x^3 x ≠ x^2

At first glance, this may seem impossible, but let's delve into the algebraic manipulations to uncover the solution.

Algebraic Manipulations

Starting with the first condition:

x x^3

Subtract x from both sides to isolate the cubic term:

x^3 - x 0

Factor out x from the left side:

x(x^2 - 1) 0

This equation implies two possible solutions:

x 0 x^2 - 1 0 which simplifies to x 1 or x -1

So, the potential solutions are x 0, x 1, and x -1.

Verification Against the Second Condition

Now, let's verify each solution against the second condition:

x 0

For x 0:

0^2 0, which is true. However, since 0^2 0, this solution does not satisfy the second condition.

x 1

For x 1:

1^2 1, which is true. However, since 1^2 1, this solution also does not satisfy the second condition.

x -1

For x -1:

-1^2 1, which is true. Since -1 ≠ 1, this solution satisfies the second condition.

Thus, the number x -1 is the only solution that meets both conditions.

Generalization for Higher Powers

Suppose we generalize the problem to x^n x, where x ≠ {0,1}. Simplifying this equation yields:

x^{n-1} 1

This equation has n-1 solutions, considering the (n-1) roots of unity. However, since x 1 does not satisfy the second condition, we lose one solution. Therefore, we are left with n-2 solutions, and one of them is x -1.

For n 3, we have:

x -1

Final Conclusion

In summary:

x^3 x but x^2 ≠ x x^3 - x 0 but x^2 - x ≠ 0 x^2 - 1 0 but x - 1 ≠ 0

Therefore, the solution that meets both conditions is:

boxed{x -1}