Proof
Let x be any real number then the statement:
P: $x^3=x^3+1$
$\frac{x^3}{x^3} = \frac{x^3}{x^3}+\frac{1}{x^3}$
$1=1+\frac{1}{x^3}$ this equation has no solutions since $\frac{1}{x^3}$ can never equal 0 and is undefined when x=0. There fore P is false.
We need to show that there exists an $x$ for which $x = x^3+1$. This is equivalent to showing that $f(x)$ has a root, where $f(x) = x^3 - x + 1$. Now use the Intermediate Value Theorem.
Question:
"Is there a number that is exactly one more than its cube?" (In my particular case, this was problem 51 from section 2.4 of Single Variable Calculus Concepts and Contexts |4e by James Stewart)
I was just assigned this problem in a homework assignment for my calculus class. I took a slightly different approach compared to @Théophile, although using the IVT here is likely the solution which most professors would look for (especially if you were asked this question on a quiz or exam where you aren't allowed a graphing calculator).
You can verify that an $x$ value which satisfies these parameters exists by graphing $y=x$ and $y=x^3+1$ as individual functions. There is one intersection between these two graphs, that gives us the $x$ value which satisfies the equation $x=(x^3)+1$
You can then plug that $x$ value into the equation $x=(x^3)+1$ to check the answer.
Note this approach serves more as an aid in conceptualizing this particular problem, rearranging the equation $x=(x^3)+1$ and using the IVT to solve for the root is a more exact means to a solution. That said, I initially found it difficult to imagine this scenario, so I thought I'd share my approach in case anyone else here is struggling in the same way.
Cheers!
