How to prove this without using Cardano's formula?

[Math100]

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?

 

[DaveE]

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.

 

[fresh_42]

$$
x^3+3x=x(x^2+3)=2^2\cdot 3^2=36
$$
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.

 

[Math100]

$$
x^3+3x=x(x^2+3)=2^2\cdot 3^2=36
$$
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?

 

[fresh_42]

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.

 

[fresh_42]

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 $0=x^2 +3x +12 \Longrightarrow x\in \left\{-\dfrac{3}{2}+\dfrac{\sqrt{-39}}{2}\, , \,-\dfrac{3}{2}-\dfrac{\sqrt{-39}}{2}\right\}.$

 

[pasmith]

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 $$\begin{split} f(x) - f(y) &= x^3 + 3x - y^3 - 3y \\ &= (x - y)(x^2 + xy + y^2 + 3)\end{split}$$ and showing that the second factor is strictly positive, which reduces to the first method on setting $y = 3$.

 

[MidgetDwarf]

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.

 

[DaveE]

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.