Three degree equation

[dduclam]

Solve the equation: [tex]x^3-x^2-x=\frac{1}{3 }[/tex] .

Very nice!

[Math Tutor]

The problem is really very interesting!

[dduclam]

The problem is really very interesting!

Thank you,teacher! Above problem have only root My solution's very nice and simple

[Guest]

x^3 - x^2 - x = 1/3

3 x^3 - 3 x^2 - 3 x = 1

3 x^3 = 3 x^2 + 3 x + 1

4 x^3 = x^3 + 3 x^2 + 3 x + 1

4 x^3 = ( x + 1)^3

Take the cube root of both sides and solve for x

[Guest]

In the following link, you will find the full derivation of the formula, together with an example of how to properly apply it
https://www.youtube.com/watch?v=zHO3YVC2T4E&list=PLfbradAXv9x4glUD2eSw384pc4zsTbzHe&index=2&ab_channel=Math%2CPhysics%2CEngineering

[Guest]

x^3 - x^2 - x = 1/3

3 x^3 - 3 x^2 - 3 x = 1

3 x^3 = 3 x^2 + 3 x + 1

4 x^3 = x^3 + 3 x^2 + 3 x + 1

4 x^3 = ( x + 1)^3

Take the cube root of both sides and solve for x

Now that's nice! Since this has been here for 7 months, I will finish it
Taking the cuber root of both sides
$4^{1/3}x= x+ 1$
$(4^{1/3}- 1)x= 1$

There are two other, non-real, roots that you can get by taking the non-real cube roots of 4 and 1.
so that $x= \frac{1}{4^{1/3}- 1}$
is the real root,