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I'm trying to figure out an example from a book with math problems. So here is an example (the equality should be proven) $$ \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}} = 4 $$

The author is suggesting to have $ \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}} = x $, so after cubing both parts of the equation we get: $$ 38 + \sqrt{1445} + 38 - \sqrt{1445} + 3\sqrt[3]{(38 + \sqrt{1445})(38 - \sqrt{1445})} \cdot x = x^3 $$ or $ x^3 + 3x - 76 = 0. $ If we substitute $4$ in the equation we'll figure out that $4$ is a root. So this part is very clear for me, but what I don't understand is the fact that the $4$ is a root of the equation is insufficient to prove the problem. The author is trying to show that the $4$ is a single root of the equation and only after that he concludes that the problem is proved.

So why is the author doing so, why proving that $4$ is a root of the equation doesn't prove the whole problem?

Michael Rozenberg
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E. Shcherbo
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    $4$ is the only $\textit {real}$ root of the cubic. And your value is clearly real. – lulu Jul 29 '19 at 14:43

3 Answers3

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The author shows that the real number $x$ is a root of the polynomial $P(x)=x^3 + 3x - 76$. Now, since $P(4)=0$, we have that $4$ is a real root of the same polynomial. Note that a polynomial of third degree could have more than one real root, but if we show that $P$ has a unique real root then we may conclude that $x$ and $4$ are the same root, that is $x=4$.

The claim is true because $$P'(x)=(x^3+3x-76)'= 3x^2+3>0$$ and therefore $P$ is a strictly increasing function which implies that the real root is unique.

Robert Z
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  • Theoretically, if this polynomial had two real roots we couldn't conclude that $x = 4$, right? If so, for me it's a bit hard to understand.. – E. Shcherbo Jul 29 '19 at 14:58
  • This is very interesting for me, I'd like to know about it more. – E. Shcherbo Jul 29 '19 at 15:01
  • Exactly. Suppose that the roots of $P$ are say $4,1,3$. Then knowing that $x$ is a root of $P$ we can only say that $x$ is one of those three numbers. – Robert Z Jul 29 '19 at 15:04
  • I got you! Thank you very much, it's a really good explanation. – E. Shcherbo Jul 29 '19 at 15:09
  • @E.Shcherbo You are welcome! – Robert Z Jul 29 '19 at 15:15
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    A simple counter example is to solve what is $2-\sqrt{16}$. Let $2-\sqrt{16}=x$. Then square both sides $4-4\sqrt{16} + \sqrt{16}^2 = x^2$. So $x^2 =4$ and ... voila! $x = 2$. But $2-\sqrt{16} = 2-4 = -2\ne 2$. We have $x^2 = 4$. And we have $2^2 = 4$ and $(-2)^2=4$. So $x$ could be either one of $2$ or $-2$. But it can't be both! – fleablood Jul 29 '19 at 15:58
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Because the substitution $\sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}}=x$ in the equation $$38 + \sqrt{1445} + 38 - \sqrt{1445}+3 \sqrt[3]{(38 + \sqrt{1445})(38 - \sqrt{1445})}\left( \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}}\right)=x^3$$ is not an equivalent transformation of the last equation.

We can get another roots except a value $\sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}}$, which we need to find.

For example, if we'll try to make the similar thing with $$1+1=x,$$ so we obtain: $$1+1+3\cdot1\cdot1(1+1)=x^3$$ and after a similar substitution $1+1=x$ in the last equation we obtain: $$2+3x=x^3$$ or $$x^3-3x-2=0$$ or $$x^3-2x^2+2x^2-4x+x-2=0$$ or $$(x-2)(x+1)^2=0,$$ which gives also $$x=-1.$$ Id est, we need to prove that the equation $$x^3+3x-76=0$$ has an unique real root, that Robert Z made.

We can use also the way, which is based on on the identity: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$ The factor $$a^2+b^2+c^2-ab-ac-bc=\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=$$ $$=\frac{1}{2}\sum_{cyc}(a^2+b^2-2ab)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$ The equality occurs for $a=b=c$ only.

In our case $a=\sqrt[3]{38 + \sqrt{1445}},$ $b=\sqrt[3]{38 - \sqrt{1445}}$ and $c=-x$ and we see that $a\neq b$, which gives $\sum\limits_{cyc}(a^2-ab)\neq0,$ which says that in our case the transform from $$a+b+c=0$$ to $$(a+b+c)\sum\limits_{cyc}(a^2-ab)=0$$ is an equivalent transformation and we don't get an extraneous root.

After this transform we obtain: $$a^3+b^3+c^3-3abc=0$$ or $$38 + \sqrt{1445} + 38 -\sqrt{1445}-x^3-3 \sqrt[3]{38 + \sqrt{1445}}\cdot\sqrt[3]{38 - \sqrt{1445}}\cdot(-x)=0$$ or $$x^3+3x-76=0$$ and since $4$ is a root of the equation, we see that this root is an unique because our transformations was equivalent.

Michael Rozenberg
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Doing so as you have writte we get $$76+2\sqrt[3]{-1}=x^3$$ and what is $$\sqrt[3]{-1}=$$?

Dr. Sonnhard Graubner
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  • $\sqrt[3]{-1} = -1$, because $ (-1)^3 = (-1) * (-1) * (-1) = -1 $ – E. Shcherbo Jul 29 '19 at 14:59
  • Ok, see here https://www.wolframalpha.com/input/?i=(-1)%5E(1%2F3) – Dr. Sonnhard Graubner Jul 29 '19 at 15:01
  • take a look, please https://math.stackexchange.com/a/25535/569639 – E. Shcherbo Jul 29 '19 at 15:04
  • @E.Shcherbo, $\sqrt[^3]{-1}=i^{2/3}$ can have imaginary roots. By Euler's Identity, we can deduct that $$i^{x}=\cos\left(\frac{\pi x}{2}\right)+i\sin\left(\frac{\pi x}{2}\right)$$Or we get $$i^{\frac{2}{3}}=\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)$$$$\sqrt[3]{-1}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$$ – Mourad Jan 28 '20 at 10:31