How to prove this without using Cardano's formula?

AI Thread Summary
The discussion revolves around proving that \( x = \sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}} = 3 \) without using Cardano's formula. The proof begins by establishing the equation \( x^3 + 3x = 36 \) and confirming that \( x = 3 \) is a solution through direct substitution. Participants explore methods to demonstrate that 3 is the only real solution, including factoring the polynomial and analyzing its derivative to show it is strictly increasing. The conversation also touches on the irreducibility of the quadratic resulting from polynomial division, reinforcing that \( x = 3 \) is the sole real root. Overall, the proof effectively combines algebraic manipulation and properties of polynomials to reach the conclusion.
Math100
Messages
813
Reaction score
229
Homework Statement
Prove without using Cardano's formula that ## \sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=3 ##.
Relevant Equations
None.
Proof:

Let ## x=\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}} ##.
Then ## x^3=(\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}})^3 ##.
Note that ## (a+b)^3=a^3+3ab(a+b)+b^3 ## where ## a=\sqrt[3]{18+\sqrt{325}} ## and ## b=\sqrt[3]{18-\sqrt{325}} ##.
This gives ## x^3=(\sqrt[3]{18+\sqrt{325}})^3+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot (\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}})+(\sqrt[3]{18-\sqrt{325}})^3\implies x^3=18+\sqrt{325}+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot x+18-\sqrt{325}\implies ##.
Now we have ## x^3=36+3\sqrt[3]{(18+\sqrt{325})(18-\sqrt{325})}\cdot x\implies x^3=36+3\sqrt[3]{-1}\cdot x\implies x^3=36-3x ##,
so ## x^3+3x=36 ##.

From here, it's clear and obvious that ## x=3 ## is a solution since ## 3^3+3(3)=36 ## by direct substitution. But how to solve this cubic equation ## x^3+3x=36 ## so that I can get ## x=3 ## to finish this proof?
 
Physics news on Phys.org
I'm not a mathematician, but I think you have finished it. If the original problem statement includes specific numbers, I don't see why you can't use those in your proof.

There are procedures to find the roots of cubic equations (ask google), but it's a lot more work than substituting a good guess into the equation.
 
$$
x^3+3x=x(x^2+3)=2^2\cdot 3^2=36
$$
We can rule out that ##x\in \{-1,+1\},## and that one or both factors are negative because of the square. So ##x\geq 2.## But ##x\geq 4## is also impossible since ##4^3=64>36.## That leaves ##x=2,3## as the only possible integer solutions and ##3^3+3\cdot 3=36.##

To solve ##x^3+3x=35## you would probably have to use Cardano or https://www.wolframalpha.com/ or use complex numbers, but that wasn't asked for.
 
Last edited:
  • Like
Likes Math100 and DaveE
fresh_42 said:
$$
x^3+3x=x(x^2+3)=2^2\cdot 3^2=36
$$
We can rule out that ##x\in \{-1,+1\},## and that one or both factors are negative because of the square. So ##x\geq 2.## But ##x\geq 4## is also impossible since ##4^3=64>36.## That leaves ##x=2,3## as the only possible integer solutions and ##3^3+3\cdot 3=36.##

To solve ##x^3+3x=35## you would probably have to use Cardano or https://www.wolframalpha.com/ or use complex numbers, but that wasn't asked for.
But how does ## x^3+3x=x(x^2+3)=2^2\cdot 3^2=36 ##? Especially where ## x(x^2+3)=2^2\cdot 3^2 ##? Also for cubic equations of the form like ## x^3+ax=b ##, we always tend to use the Cardano's formula? There's no other formula?
 
Math100 said:
But how does ## x^3+3x=x(x^2+3)=2^2\cdot 3^2=36 ##? Especially where ## x(x^2+3)=2^2\cdot 3^2 ##? Also for cubic equations of the form like ## x^3+ax=b ##, we always tend to use the Cardano's formula? There's no other formula?
Assuming we are looking for an integer solution.

I first thought that the prime factors would lead to a good answer. A prime number ##p## is a number that isn't invertible, i.e. that isn't ##\pm 1## in the case of integers, and for which holds that ##p\,|\,a\cdot b## implies ##p\,|\,a ## or ##p\,|\,b.##

My idea was: ##36=x^3+3x=x\cdot(x^2+3)## means that ##3\,|\,x\cdot(x^2+3)## and so that ##3\,|\,x## or ##3\,|\,(x^2+3).## If ## 3\,|\,x, ## i.e. ##x=3k## then ##x^2+3=9k^2+3 \,|\,36 ## only if ##k=1## meaning ##x=3.## Since ##x=3## is a solution we are done.

Of course, there are always two additional complex solutions:
https://www.wolframalpha.com/input?i=x^3+3x=36

These considerations used properties of the integer numbers. So if there was only a non-integer real solution, things would have been different.
 
Last edited:
Here is another idea:
$$
x^3+3x=36
$$
We can find or guess the solution ##x=3.## If ##x^3+3x=36## has an integer solution, then it has to be a divisor of ##36.## ##x\geq 4## are too big, ##x\leq 2## are too small. So ##x=3## is the only remaining. Now,
$$
(x^3+3x-36)\, : \,(x-3)\, = \,x^2 +3x +12
$$
We get the remaining roots by ##0=x^2 +3x +12 \Longrightarrow x\in \left\{-\dfrac{3}{2}+\dfrac{\sqrt{-39}}{2}\, , \,-\dfrac{3}{2}-\dfrac{\sqrt{-39}}{2}\right\}.##
 
  • Like
Likes DaveE and Math100
Math100 said:
Homework Statement: Prove without using Cardano's formula that ## \sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}}=3 ##.
Relevant Equations: None.

Proof:

Let ## x=\sqrt[3]{18+\sqrt{325}}+\sqrt[3]{18-\sqrt{325}} ##.

[...]

so ## x^3+3x=36 ##.

From here, it's clear and obvious that ## x=3 ## is a solution since ## 3^3+3(3)=36 ## by direct substitution. But how to solve this cubic equation ## x^3+3x=36 ## so that I can get ## x=3 ## to finish this proof?

You have shown that x = 3 is a real solution. All that remains is to show that is it the only real solution. To that end, you can write x^3 + 3x - 36 = (x- 3)\left((x - a)^2 + b\right). Expanding the right hand side and comparing coefficients of powers of x will give you a and b.

Alternatively, it should be fairly obvious that the derivative of f(x) = x^3 + 3x - 36 is strictly positive; hence f has at most one real root. If you don't want to use calculus, you can still show that it is strictly increasing by factorizing <br /> \begin{split}<br /> f(x) - f(y) &amp;= x^3 + 3x - y^3 - 3y \\<br /> &amp;= (x - y)(x^2 + xy + y^2 + 3)\end{split} and showing that the second factor is strictly positive, which reduces to the first method on setting y = 3.
 
Last edited:
I am assuming since this is posted in precal section, it safe to assume we are working with a field., ie., R?

If so, once op arrives at the polynomial. If we plug in 3 into the polynomial we get a zero.

Therefor x-3 is a linear factor.

Then we use divide the polynomial by the linear factor, and find out what the quadratic is [post 6 by fresh].

If we are assuming a field, then we can use Eisenstein's Criterion with p=3. Then it would follow that the quadratic is irreducible which implies that the quadratic has no linear factors.

Therefore x = 3 is the only real root.

I may be wrong [its been a while since I did ring theory], and naturally, I would proceed with the Calculus approach as mentioned above.
 
MidgetDwarf said:
I am assuming since this is posted in precal section, it safe to assume we are working with a field., ie., R?
I learned about complex numbers and polynomial roots a few years before I knew any calculus. I don't even really see much connection between the two subjects.
 
Back
Top