Question

Fill in the blanks, to complete the factorization:


(a^2 + b^2 - c^2)^2 - 4a^2 b^2 - 4a^2 c^2 + 4b^2 c^2 = (a + ___)(a + ___)(a + ___)(a + ___)

Answer 1

To complete the factorization of the expression

\[
(a^2 + b^2 - c^2)^2 - 4a^2 b^2 - 4a^2 c^2 + 4b^2 c^2,
\]

we can simplify it step by step.

Let's first recognize that the expression can be potentially factored by rearranging or regrouping it. It takes a form that can be examined for known identities or factorable patterns.

We can rewrite the expression:

\[
(a^2 + b^2 - c^2)^2 - (2ab)^2 - (2ac)^2 + (2bc)^2.
\]

Recognizing this resembles a squared term minus a sum of squared terms, we can see if it matches any known identities.

After simplification and manipulation, one can check that the expression yields:

\[
(a + b - c)^2 (a - b + c)^2.
\]

Thus we find that it can indeed be expressed in terms of products of simpler binomials.

Now, to write down the correct factorization in the specified form:

The expression can be factored as:

\[
(a + b - c)(a + b - c)(a - b + c)(a - b + c).
\]

So, filling in the blanks in the requested format gives us:

\[
(a + b - c)(a + b - c)(a - b + c)(a - b + c).
\]

Therefore, the filled-in blanks are:

\( b - c \) in the first and second blanks, and \( -b + c \) in the third and fourth blanks.

Thus, the full factorization expression to fill the blanks is:

\[
(a + b - c)(a + b - c)(a - b + c)(a - b + c).
\]

Final answer:

\[
(b - c), (b - c), (-b + c), (-b + c).
\]