Prove equality - Necessity and sufficiency 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?
 A: 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.                   
A: 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.
A: Doing so as you have writte we get
$$76+2\sqrt[3]{-1}=x^3$$ and what is $$\sqrt[3]{-1}=$$?
