Irrational Equation

[Math Tutor]

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

[Guest]

This question is solvable.

If you cube it first, it comes out as 3x + 6 = 0
This means that 3x = -6
This means that x = -2

Which if you place into the equation it comes out as (Cube root of -1) + (Cube root of 0) + (Cube root of 1) = 0. Which is -1 + 0 + 1 = 0

[Guest]

How do you prove that there are no other roots?

[Guest]

The previous post is nonsense (though does give the right answer).

Take the middle term to the other side to get
[tex]\sqrt[3]{x+1}+\sqrt[3]{x+3}=-\sqrt[3]{x+2}[/tex]
Cube both sides to get
[tex](x+1)+3\sqrt[3]{(x+1)^2(x+3)}+3\sqrt[3]{(x+1)(x+3)^2}+(x+3)=-(x+2)[/tex]
which rearranges to
[tex]3\sqrt[3]{(x+1)^2(x+3)}+3\sqrt[3]{(x+1)(x+3)^2}=-3(x+2)[/tex]
"factorizing" the left hand side gives
[tex]3\sqrt[3]{(x+1)(x+3)}(\sqrt[3]{x+1}+\sqrt[3]{x+3})=-3(x+2)[/tex]
Looking back to the very first statement I made, we see that we can substitute it into the left hand side
[tex]3\sqrt[3]{(x+1)(x+3)}(-\sqrt[3]{x+2})=-3(x+2)[/tex]
Cubing both sides gets us
[tex]-27(x+1)(x+3)(x+2)=-27(x+2)^3[/tex]
Now it is easy to see that either x+2=0 (giving the solution x=-2) or we can divide both sides by (x+2)
The latter case leads to
[tex]-27(x+1)(x+3)=-27(x+2)^2[/tex]
which has no solutions.

So the only solution is x=-2.

Hope that helped

R. Baber.

[Math Tutor]

Baber, your solution is very good.