Why substitution in irrational equation doesn't give equivalent equation?

Question

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.

Answer 1

Very good question! The first example is misleading in writing equivalences everywhere; the second equivalence $$3-x+ 9\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x})+6+x$$ $$\iff$$ $$3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27$$ should (at least conceptually) be an implication $(\implies)$. It is true that it is in fact an equivalence, but this is not yet clear at this point. Let me explain:

The argument starts from the assumption that if $x$ satisfies $$\sqrt[3]{3-x} + \sqrt[3]{6+x}=3,$$ then it also satisfies the expressions that follow. In both examples the first equivalence stems from the simple fact that $$x^3=y^3\quad\iff\quad x=y,$$ but the second equivalence uses a substitution that need not be reversible; a number $x$ can satisfy $$3 \sqrt[3]{(x+1)(3x+1)(x-1)} = -3x-3,$$ but this does not imply that it should satisfy $$\sqrt[3]{x+1} + \sqrt[3]{3x+1} = \sqrt[3]{x-1}.$$ This is illustrated by the solution $x=0$. In the first example it just so happens that we do not get any extra solutions, and hence, with hindsight, this implication turns out to be an equivalence. But that is not at all clear before checking whether all solutions to the last equation are also solutions to the original equation.

So to answer your question; yes, you should check whether all solutions to the last equation are also solutions to the original equation. If you are certain that every step along the way is unambiguously reversible, then you do not need to check. But if it is not clear whether a step is reversible (as in both examples here), you should check.

Answer 2

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.

1. $$\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$$ 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: $$\{-1\}$$