Solve x⁴ - 16 = (x-2)²(x+2)²: Polynomial Identity Verification

First, let's examine the given equation: $x^4 - 16 = (x-2)(x-2)(2+x)(2+x)$.

The expression on the right-hand side can be rewritten and simplified by recognizing it as a repeated multiplication form:

$(x-2)^2(2+x)^2 = [(x-2)(2+x)]^2$.

Now, apply the difference of squares to the product $(x-2)(2+x)$:

$(x-2)(2+x) = x^2 - 4$.

Now, squaring $x^2 - 4$ gives:

$(x^2 - 4)^2 = x^4 - 8x^2 + 16$.

Compare this to the left side of the equation $x^4 - 16$.

The terms do not match, indicating $x^4 - 16$ does not equal $x^4 - 8x^2 + 16$. Thus, originally factoring may contain inherent assumption errors or unrealistic roots if said polynomial equation was proposed as true for all $x$.

However, for the equality $x^4 - 16 = 0$ specifically, solving this separately would involve:

Recognizing the equation as a difference of squares:

$(x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4)$.

Factor further: $(x^2 - 2^2) = (x-2)(x+2)$.

The $x^2 + 4$ does not factor into real numbers as it results in imaginary roots, thus focus remains on the two real roots of the quadratic factor:

$x = \pm 2$.

Therefore, the valid real solutions are found to be $\pm 2$.

Thus, the final correct solution is: $\pm 2$.