To determine if there exists a number that is exactly 3 more than its cube, we can formulate the problem mathematically. Let’s denote the number as x. According to the problem, we are looking for values of x that satisfy the equation:
x^3 – x – 3 = 0
This equation rearranges to indicate that we’re seeking the roots of a polynomial equation. We will evaluate potential solutions either through substitution or numerical methods.
1. Testing Possible Solutions
Firstly, let’s test a few small integer values to see if they yield a solution:
- For x = 1:
- For x = 2:
- For x = 3:
- For x = 0:
- For x = -1:
- For x = -2:
1^3 – 1 – 3 = 1 – 1 – 3 = -3 (not a solution)
2^3 – 2 – 3 = 8 – 2 – 3 = 3 (not a solution)
3^3 – 3 – 3 = 27 – 3 – 3 = 21 (not a solution)
0^3 – 0 – 3 = 0 – 0 – 3 = -3 (not a solution)
-1^3 – -1 – 3 = -1 + 1 – 3 = -3 (not a solution)
-2^3 – -2 – 3 = -8 + 2 – 3 = -9 (not a solution)
2. Checking for Non-Integer Solutions
Since none of the integer values worked, let’s check for non-integer values. The function f(x) = x^3 – x – 3 is continuous, and we can use methods like the Intermediate Value Theorem to search for roots.
By evaluating the function at x = 1.5:
1.5^3 – 1.5 – 3 = 3.375 – 1.5 – 3 = -1.125 (which is less than 0)
Now, at x = 2 (as checked earlier):
8 – 2 – 3 = 3 (which is greater than 0)
This indicates that there is a root between 1.5 and 2. Using numerical methods like the bisection method or Newton’s method could help pinpoint the root more accurately. If you apply these methods, you would find a number around 1.8793852415718. This approximate value solves the equation, showing that indeed, there is a number that fits the criteria you are looking for.
Conclusion
In conclusion, yes, there is a number that is exactly 3 more than its cube. Specifically, the approximate value of this number is about 1.8794.