Solving the Equation (x^2x+1)^4+4x^2(x^2x+1)^2=5x^4
This equation looks complicated, but we can solve it by using some clever algebraic manipulations.
Making a Substitution
Let's start by making a substitution to simplify the equation. Let y = x^2  x + 1. Now our equation becomes:
y^4 + 4x^2y^2 = 5x^4
Rearranging and Factoring
We can rearrange this equation to look like a quadratic:
y^4 + 4x^2y^2  5x^4 = 0
Now we can factor the lefthand side:
(y^2 + 5x^2)(y^2  x^2) = 0
Solving for y
This gives us two possible equations:
 y^2 + 5x^2 = 0
 y^2  x^2 = 0
Let's analyze each equation:

Equation 1: y^2 + 5x^2 = 0. Since both y^2 and 5x^2 are always nonnegative, the only way this equation can hold is if both y^2 and 5x^2 are equal to zero. This means y = 0 and x = 0.

Equation 2: y^2  x^2 = 0. This can be factored as (y + x)(y  x) = 0. This leads to two possibilities:
 y + x = 0 => y = x
 y  x = 0 => y = x
Substituting Back and Solving for x
Now we need to substitute back our original value for y (y = x^2  x + 1) and solve for x:

Case 1: y = 0
 x^2  x + 1 = 0. This quadratic equation has no real solutions.

Case 2: y = x
 x^2  x + 1 = x. This simplifies to x^2 + 1 = 0, which has no real solutions.

Case 3: y = x
 x^2  x + 1 = x. This simplifies to x^2  2x + 1 = 0. Factoring this gives us (x  1)^2 = 0, leading to x = 1.
Conclusion
The only real solution to the equation (x^2x+1)^4+4x^2(x^2x+1)^2=5x^4 is x = 1.