<s>[INST] Why substitution in irrational equation doesn't give equivalent equation? I have two examples of irrational equations:
The first example: $\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$
In solution, they take cube of both sides and do following:
\begin{eqnarray*}
&\sqrt[3]{3-x} &+ \sqrt[3]{6+x}=3\\
&\iff&
3-x+ 3\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x}) +6+x=27\\
&\iff&
3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27   \iff \sqrt[3]{(3-x)(6+x)} = 2\\
&\iff&
x^2+3x-10=0\\
&\iff&   x=2\quad  \text{ or }\quad x= -5
\end{eqnarray*}
They conclude that both values are solutions, they satisfy the original equation.
The second example: $\sqrt[3]{x+1} +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}$
Here they do the following:
\begin{eqnarray*}
&\sqrt[3]{x+1}& +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}\\
&\iff&
x+1 + 3 \sqrt[3]{(x+1)(3x+1)} (\sqrt[3]{x+1} +  \sqrt[3]{3x+1} ) + 3x+1 = x-1\\
&\implies&
3 \sqrt[3]{(x+1)(3x+1)(x-1)} = -3x-3\\
&\iff&
(x^2-1)(3x+1)= -(x+1)^3\\
&\iff&
x= 0 \quad\text{ or }\quad x = -1
\end{eqnarray*}
but the only solutions is $x= -1$, because $0$ doesn't satisfy the equation.

What is difference between these examples, why do we have all equivalence signs in the first, and implication in the secondnd example?

The explanation in solution is that we have substituted  $\sqrt[3]{3-x} + \sqrt[3]{6+x}$ by a number in the 1st example, and in the 2nd example we substituted by another expression which depends on x. I don't understand that, when does substitution  give equivalent equation? Do we always have to verify if solutions satisfy the original equation in the end (is that the answer)?
Please help with this explanation, I need to understand better the irrational equations. Thanks a lot in advance.
 [/INST] These problems don't happen if you'll use the following identity.
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Since, $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2+b^2-2ab)=\frac{1}{2}\sum_{cyc}(a-b)^2,$$
we see that $\sum\limits_{cyc}(a-b)^2=0$ for $a=b=c$ only and it can give a extraneous root of the equation.
Now, we can solve your equations by using this idea.

*

*$$\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$$
Here $a=\sqrt[3]{3-x},$ $b=\sqrt[3]{6+x}$ and $c=-3.$
Since $$\sqrt[3]{3-x}=\sqrt[3]{6+x}=-3$$ is impossible, our equation is equivalent to:
$$3-x+6+x-27-3\sqrt[3]{(3-x)(6+x)}(-3)=0$$ or
$$\sqrt[3]{(3-x)(6+x)}=-2,$$ which gives the answer:
$$\{2,-5\}$$
2. $$\sqrt[3]{x+1} +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}.$$
Since $$x+1=3x+1=-(x+1)$$ is possible for $x=0$ and $0$ is not a root of our equation, we need to remove this root before we'll write the answer.
Id est, we obtain:
$$x+1+3x+1-(x-1)+3\sqrt[3]{(x+1)(3x+1)(x-1)}=0$$ or
$$\sqrt[3]{(x+1)(3x+1)(x-1)}=-x-1$$ or
$$x^2(x+1)=0,$$ which gives the answer:
$$\{-1\}$$
</s>