How to prove this without using Cardano's formula?

  • #1
Math100
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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?
 
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  • #2
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.
 
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  • #3
$$
x^3+3x=x(x^2+3)=2^2\cdot 3^2=36
$$
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.
 
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  • #4
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?
 
  • #5
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.
 
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  • #6
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\}.##
 
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  • #7
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 [itex]x = 3[/itex] is a real solution. All that remains is to show that is it the only real solution. To that end, you can write [tex]x^3 + 3x - 36 = (x- 3)\left((x - a)^2 + b\right).[/tex] Expanding the right hand side and comparing coefficients of powers of [itex]x[/itex] will give you [itex]a[/itex] and [itex]b[/itex].

Alternatively, it should be fairly obvious that the derivative of [itex]f(x) = x^3 + 3x - 36[/itex] is strictly positive; hence [itex]f[/itex] 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 [tex]
\begin{split}
f(x) - f(y) &= x^3 + 3x - y^3 - 3y \\
&= (x - y)(x^2 + xy + y^2 + 3)\end{split}[/tex] and showing that the second factor is strictly positive, which reduces to the first method on setting [itex]y = 3[/itex].
 
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  • #8
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.
 
  • #9
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.
 
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