Solving the Equation x3 x3 x6: A Comprehensive Guide

Can the equation x3 x3 x6 be true? Let's dive into the algebraic steps and explore the solutions for this equation. This guide will walk you through the process of simplifying, factoring, and solving the equation, highlighting key algebraic principles.

Simplifying the Equation

Let's start by simplifying the left side of the equation:

x3 x3 can be simplified to 2x3.

So, the equation becomes: 2x3 x6.

Rearrange the equation to: x6 - 2x3 0.

Factoring the Equation

Next, we factor out x3 from the equation:

x6 - 2x3 0 can be written as: x3(x3 - 2) 0.

This gives us two factors to consider: x3 0 and x3 - 2 0.

Solving the Factors

Let's solve these factors one by one:

x3 0

This equation simplifies to x 0.

Substituting x 0 into the original equation, we get:

03 03 06

This simplifies to 0 0, which is true.

x3 - 2 0

Adding 2 to both sides gives us x3 2.

Taking the cube root of both sides, we get:

x 21/3

This value can be either a real number or a complex number (roots).

Algebraic Roots of the Polynomial

To find all the roots of the equation, we can use the approach of solving the polynomial x6 - 2x3 0. Let's set y x3 and solve for y2 - 2y 0 first:

This gives us y 0 or y 2.

Substituting back, we get x3 0 or x3 2.

Hence, the solutions for x3 0 are x 0 (with multiplicity three).

The solution for x3 2 is x 21/3, which can have three roots: one real and two complex.

So, the complete set of solutions for the equation x3 x3 x6 includes:

x 0 (three roots)

x 21/3

The two complex roots derived from the form x frac{-21/3 pm isqrt{3 cdot 22/3}}{2}.

Conclusion

The equation x3 x3 x6 holds true for the following values of x:

x 0 (with multiplicity three)

x 21/3

The two complex roots: frac{-21/3 pm isqrt{3 cdot 22/3}}{2}

These results showcase the power of algebraic manipulation and factoring in solving polynomial equations, providing a thorough insight into the nature of the solutions.