Stuck with a complicated step in an equation

[gamaz321]

Hello,
I am trying to solve an equation. However, it is getting too complicated to handle after a few steps. I would appreciate any help to allow me to proceed further.
Thank you.

 

[skeeter]

note [imath] (x^3+ 1)^3 = x^9 + 3x^6 + 3x^3 + 1[/imath]

[imath]x^9+3x^6+3x^3-16x+9 = 0[/imath]

by observation, note [imath]x=1[/imath] is a solution

 

[BigBeachBanana]

note [imath] (x^3+ 1)^3 = x^9 + 3x^6 + 3x^3 + 1[/imath]

[imath]x^9+3x^6+3x^3-16x+9 = 0[/imath]

by observation, note [imath]x=1[/imath] is a solution

@gamaz321 That's only one solution. There's more.

 

[BigBeachBanana]

Hello,
I am trying to solve an equation. However, it is getting too complicated to handle after a few steps. I would appreciate any help to allow me to proceed further.
Thank you.

It's a tough problem and difficult to solve through ordinary methods.

Observe [imath]\frac{x^3+1}{2}[/imath] and [imath]\sqrt[3]{2x-1}[/imath] are inverses of each other.

Can you continue with this hint?

 

[skeeter]

@gamaz321 That's only one solution. There's more.

that’s why I said x = 1 is a solution

 

[gamaz321]

It's a tough problem and difficult to solve through ordinary methods.

Observe [imath]\frac{x^3+1}{2}[/imath] and [imath]\sqrt[3]{2x-1}[/imath] are inverses of each other.

Can you continue with this hint?

I could not proceed even with the suggestion. It appears too complex. Sorry. Thanks anyways.

 

[BigBeachBanana]

I could not proceed even with the suggestion. It appears too complex. Sorry. Thanks anyways.

It's less complicated than you think, and the solution is quite elegant. Here's another hint.
The function is reflected over the line [imath]y=x[/imath] to obtain the inverse. What does this tell you? If you still have trouble, graph all of the functions.

 

[gamaz321]

Ok, thanks I am going to try again with the new hint.

 

[BigBeachBanana]

Not sure what happened with the OP, but this was an interesting problem.
By observing that [imath]\frac{x^3+1}{2}[/imath] and [imath]\sqrt[3]{2x-1}[/imath] are inverses of each other, this implies that they intersect on the line [imath]y=x.[/imath]
Then the solution can be found by [math]\frac{x^3+1}{2}=x\implies x^3-2x+1=(x-1)(x^2+x-1)=0\implies x=1,\frac{-1 \pm \sqrt{5}}{2}[/math]

 

[gamaz321]

Thanks for the nice solution. I appreciate it!