(2x^2-2x-1)^3+(2x-1)^3=(x^2-x+1)^3+(x^2+x-3)^3

4 min read Jun 16, 2024

Solving the Equation: (2x^2-2x-1)^3+(2x-1)^3=(x^2-x+1)^3+(x^2+x-3)^3

This equation presents an interesting challenge due to the high powers involved. However, we can solve it by employing a clever algebraic trick and factoring. Let's break down the solution step-by-step:

1. Recognizing the Pattern

Observe that the equation resembles the expansion of the following identity:

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

We can rewrite our equation by grouping terms:

[(2x^2-2x-1)^3 + (2x-1)^3] = [(x^2-x+1)^3 + (x^2+x-3)^3]

2. Applying the Identity

Now, we can apply the cubic identity to both sides of the equation. Let:

a = 2x^2 - 2x - 1
b = 2x - 1
c = x^2 - x + 1
d = x^2 + x - 3

This gives us:

[(a + b)^3] = [(c + d)^3]

Expanding the cubes:

a^3 + 3a^2b + 3ab^2 + b^3 = c^3 + 3c^2d + 3cd^2 + d^3

3. Simplifying and Factoring

Now, substitute back the original expressions for a, b, c, and d, and simplify the equation. You'll notice that many terms cancel out, leaving us with:

3(2x^2-2x-1)^2(2x-1) + 3(2x^2-2x-1)(2x-1)^2 = 3(x^2-x+1)^2(x^2+x-3) + 3(x^2-x+1)(x^2+x-3)^2

We can factor out a 3 from both sides and further simplify to:

(2x^2-2x-1)(2x-1)[(2x^2-2x-1) + (2x-1)] = (x^2-x+1)(x^2+x-3)[(x^2-x+1) + (x^2+x-3)]

Simplifying further:

(2x^2-2x-1)(2x-1)(2x^2-1) = (x^2-x+1)(x^2+x-3)(2x^2-2)

4. Solving the Equation

Now, we have a simpler equation that we can solve. Observe that both sides have a common factor of 2:

(2x^2-2x-1)(2x-1)(2x^2-1) = 2(x^2-x+1)(x^2+x-3)(x^2-1)

Since we have a product equal to zero, at least one of the factors must be equal to zero. This leads to several possible solutions:

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

Solve each of these quadratic equations using the quadratic formula or factoring techniques to obtain the final solutions for x.

Note: Remember to verify your solutions by substituting them back into the original equation. Some solutions might be extraneous, meaning they don't satisfy the original equation.