Solving equation with cube roots.

[chingel]

Homework Statement

Solve the equation:

$\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0$

The Attempt at a Solution

What I did was move (x+3)^(1/3) to the other side, cube both sides and when I put them equal to 0 again, I managed to factor (x+2)^(1/3) out of it giving one solution x=-2.

However if I looked up the proposed solution after cubing and a little gathering and grouping it arrived at this:

$3x+6=-3\sqrt[3]{x+1}*\sqrt[3]{x+2}*(\sqrt[3]{x+1}+\sqrt[3]{x+2})$

Which is clear, but then in the next step it is converted to this without explanation:

$x+2=\sqrt[3]{(x+1)(x+2)(x+3)}$

From there on the solution is just putting it equal to 0 and factorizing, but how did it get to this from the previous? It would imply $-(\sqrt[3]{x+1}+\sqrt[3]{x+2})$ is equal to $\sqrt[3]{(x+3)}$, which I don't think it is. Is it wrong or what is up with that?

 

[phyzguy]

Look at your original equation:
$$\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0$$
This means:
$$\sqrt[3]{x+1}+\sqrt[3]{x+2}=-\sqrt[3]{x+3}$$
Plug this into:
$$3x+6=-3\sqrt[3]{x+1}*\sqrt[3]{x+2}*(\sqrt[3]{x+1}+\sqrt[3]{x+2})$$
and cancel out a factor of 3 to give:
$$x+2=\sqrt[3]{(x+1)(x+2)(x+3)}$$

 

[chingel]

Oh, didn't see that, thank you.