Finding roots of the equation $2 \sqrt[3]{2x-1}=x^3+1$,

Question

The Equation is

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

I divided by $2$ and then cubed both sides,

$$2x-1=\frac{x^9+3x^6+3x^3+1}{8}$$

Now, if I make the substitution $y=x^3$ the equation becomes more complicated.

I need help to proceed from here. Thanks

You would then need to put $x=y^{1/3}$ into $2x=1$ portion, so yes would get more complicated/ — coffeemath, Mar 08 '20 at 12:27
Welcome to Mathematics Stack Exchange. By inspection, $x=1$ is a solution — J. W. Tanner, Mar 08 '20 at 12:30

Dietrich Burde · Answer 1 · 2020-03-08T12:59:08.360

2

You have obtained the equation $$ x^9 + 3x^6 + 3x^3 - 16x + 9=0. $$ By the rational root theorem we see that $x=1$ is a root. Furthermore we can factorize the equation as $$ (x^6 + 2x^4 + 2x^3 + 4x^2 + 2x + 9)(x^2 + x - 1)(x - 1)=0 $$ and the polynomial of degree $6$ has no real roots (why?). So the only solutions are $x=1$ and the two solutions from the quadratic equation.

edited Mar 08 '20 at 12:59

answered Mar 08 '20 at 12:28

Dietrich Burde

140,055

2

Why are there no roots for the sixth degree factor? – Gkopi Mar 08 '20 at 12:37
@Gkopi There are indeed six roots, but none of them is real. See for example here. – Dietrich Burde Mar 08 '20 at 12:56

score 0 · Accepted Answer · answered Mar 08 '20 at 12:30

Rewrite it as

$\sqrt[3]{2x-1}=\frac{x^3+1}{2}$

Now let $y=\sqrt[3]{2x-1}$

$y=\frac{x^3+1}{2}$

Now,

This can be rewritten as

$x=\sqrt[3]{2y-1}$

If $x=f(y)$, $y=f(x)$

We can conclude that

$$x=y$$

Now,

$$x^3-2x+1=0$$

And this cubic is easy

$$y=1,\frac{-1\pm 5}{2}$$

Finding roots of the equation $2 \sqrt[3]{2x-1}=x^3+1$,

2 Answers2