How to Solve the Equation (2 sqrt[3]{2x-1} x^3 1)

This article will take you through the detailed steps to solve a complex equation: (2 sqrt[3]{2x-1} x^3 1). By following the methods and techniques outlined below, you will understand the process from isolating the cube root to finding all possible solutions, including the use of numerical methods and polynomial analysis.

Step-by-Step Solution

To solve the equation 2 sqrt[3]{2x-1} x^3 1, we start by isolating the cube root.

Isolating the Cube Root

First, we divide both sides by 2:

[sqrt[3]{2x-1} frac{x^3 1}{2}]

Eliminating the Cube Root

Next, we cube both sides to eliminate the cube root:

[2x - 1 left(frac{x^3 1}{2}right)^3]

Expanding the Right Side

Expanding the right side gives us:

[left(frac{x^3 1}{2}right)^3 frac{(x^3 1)^3}{8}]

So the equation becomes:

[2x - 1 frac{(x^3 1)^3}{8}]

Eliminating the Fraction

To eliminate the fraction, we multiply both sides by 8:

[16x - 8 (x^3 1)^3]

This results in a polynomial equation:

[x^3 1^3 - 16x - 8 0]

Checking for Rational Roots

The equation can be quite complex to solve directly. To find possible rational roots, we can use the Rational Root Theorem. Let's test some simple values for x.

Testing Simple Values

First, try x 2:

[2^3 1^3 - 16 cdot 2 - 8 8 1 - 32 - 8 9 - 32 - 8 -21 eq 0]

Next, try x 1:

[1^3 1^3 - 16 cdot 1 - 8 1 1 - 16 - 8 2 - 24 -22 eq 0]

After testing these values, we find that x 1 is a root. We can factor x - 1 out of the polynomial.

Factoring and Finding Other Roots

Using synthetic division, we can divide x^3 1^3 - 16x - 8 by x - 1.

Result

From these initial tests, we have found that x 1 is a solution. Further roots can be examined with appropriate methods such as numerical methods or graphing.

Addendum: Discussion of Additional Roots by Jack Smith

Jack Smith provides additional solutions, stating that:

[x 1, frac{-1 pm sqrt{5}}{2}]

Let a frac{-1 pm sqrt{5}}{2}. Then it can be shown that:

[a^3 - 1^3 82a - 1]

By taking cube roots, we see that (x a) is a solution. This leaves the problem of showing that:

[f(x) x^6 - 2x^4 - 2x^3 - 4x^2 - 2x - 9]

has no real zeros.

Proving (f(x) geq 0)

Jack uses the polynomial:

[g(x) x^4 - x^3 - 2x^2 - x - 4.5]

We will prove that g(x) geq 0 for x geq 0.

For x geq 1

If x geq 1, then:

[g(x) x^3 - x cdot x - x^2 - 4.5 geq 0 - 1 - 4.5 -5.5 geq 0]

For 0 leq x leq 1

If 0 leq x leq 1, then:

[g(x) x^3 - 2x cdot x - x^2 - 4.5 leq 0 - 2 cdot 1 - 1 - 4.5 -8.5 leq 0]

Thus, g(x) geq 0 for x geq 0. Clearly, f(x) geq 0 for x geq 0. If x , then f(x) x^6 - 2g(-x) geq 0 by the previous result.