Solving the Cubic Equation: (x^2+3x4)^3+(2x^25x+3)^3=(3x^22x1)^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 higherorder 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.