(x^2+3x-4)^3+(2x^2-5x+3)^3=(3x^2-2x-1)^3

4 min read Jun 17, 2024

Solving the Cubic Equation: (x^2+3x-4)^3+(2x^2-5x+3)^3=(3x^2-2x-1)^3

This equation appears complex, but we can solve it using a clever algebraic trick and some basic factorization. Here's how:

Applying the Sum of Cubes Formula

We can recognize that the equation is in the form of the sum of cubes: a³ + b³ = (a + b)(a² - ab + b²)

Let's define our 'a' and 'b' terms:

a = x² + 3x - 4
b = 2x² - 5x + 3

Substituting these into the sum of cubes formula, we get:

(x² + 3x - 4 + 2x² - 5x + 3)[(x² + 3x - 4)² - (x² + 3x - 4)(2x² - 5x + 3) + (2x² - 5x + 3)²] = (3x² - 2x - 1)³

Simplifying the Equation

Now, let's simplify the equation step by step:

Combine the terms inside the first bracket: (3x² - 2x - 1)[(x² + 3x - 4)² - (x² + 3x - 4)(2x² - 5x + 3) + (2x² - 5x + 3)²] = (3x² - 2x - 1)³
Notice that the expression in the second bracket is also in the form of the sum of cubes! Let's define:
- c = x² + 3x - 4
- d = 2x² - 5x + 3
This allows us to rewrite the equation as:

(3x² - 2x - 1)[(c + d)(c² - cd + d²)] = (3x² - 2x - 1)³
Simplify further:

(3x² - 2x - 1)[(3x² - 2x - 1)(c² - cd + d²)] = (3x² - 2x - 1)³

Finding the Solutions

We have now reached a point where we can easily solve the equation. Notice that:

(3x² - 2x - 1) is a common factor on both sides of the equation.

Therefore, we have two possible scenarios:

(3x² - 2x - 1) = 0 This quadratic equation can be solved using the quadratic formula or factoring. It yields two solutions for x.
(c² - cd + d²) = (3x² - 2x - 1)² This equation is more complex, but it represents the remaining solutions. We can substitute back the values of 'c' and 'd' and expand the equation. However, this will likely lead to a higher-order polynomial equation that might be difficult to solve directly.

Conclusion

In conclusion, solving the equation (x² + 3x - 4)³ + (2x² - 5x + 3)³ = (3x² - 2x - 1)³ involves a clever application of the sum of cubes formula and simplification. It leads to two possible scenarios, resulting in two straightforward solutions from the quadratic equation and potentially more complex solutions from the remaining equation.