Factorise:
a^4 + b^4 + c^4 - 2 (a^2 x b^2 + a^2 x c + b^2 x c)
To factorize the given expression:
a^4 + b^4 + c^4 - 2(a^2 * b^2 + a^2 * c + b^2 * c)
We can observe that this expression is in the form of a perfect square. Let's consider each term separately and try to rewrite them in a more factorizable form.
First, let's look at a^4 + b^4 + c^4.
This expression can be factorized using the sum of cubes formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2).
So, we can rewrite a^4 + b^4 + c^4 as (a^2)^2 + (b^2)^2 + (c^2)^2.
Now, let's look at the second term: -2(a^2 * b^2 + a^2 * c + b^2 * c).
We can factor out a common factor of 2 from each term: -2(a^2 * b^2 + a^2 * c + b^2 * c) = 2(-a^2 * b^2 - a^2 * c - b^2 * c).
Now, if we group the negative terms and factor out (-1), we get: 2[-(a^2 * b^2 + a^2 * c + b^2 * c)] = 2[-(a^2 * b^2 + a^2 * c + b^2 * c)] = -2(a^2 * b^2 + a^2 * c + b^2 * c).
Now, let's factorize the negative terms:
-2(a^2 * b^2 + a^2 * c + b^2 * c) = -2[a^2 * (b^2 + c) + b^2 * c].
Now, putting everything together, we have:
a^4 + b^4 + c^4 - 2(a^2 * b^2 + a^2 * c + b^2 * c) = (a^2)^2 + (b^2)^2 + (c^2)^2 - 2[a^2 * (b^2 + c) + b^2 * c].
This expression is now in a factorizable form.