To prove that the given expression is a perfect square, we need to demonstrate that it can be written in the form (x + y)², where x and y are real numbers.



Let's expand the given expression:
-a² ab + ac + ab + a - b² bc + ac + bc - c²

Simplifying, we have:
-a² ab + ab + a + ac + ac - b² bc + bc - c²

Rearranging the terms, we get:
-ab(a + 1) + ac(a + 1) - bc(b² - 1) + bc - c²

Factoring out common terms, we obtain:
(ab - ac)(a + 1) + bc(1 - b²) + bc - c²

Now, let's simplify this expression further.



(ab - ac)(a + 1) + bc(1 - b²) + bc - c²

Using the identity (a - b)(a + b) = a² - b², we can rewrite the expression as:
(ab - ac)(a + 1) + bc(1 - b)(1 + b) + bc - c²

Applying the distributive property, we can further simplify:
(ab - ac)(a + 1) + bc(1 - b)(1 + b) + bc - c²
(ab - ac)(a + 1) + bc(1 - b)(1 + b) + (bc - c²)

Now, let's focus on the second term: bc(1 - b)(1 + b). Expanding this expression, we have:
bc(1 - b)(1 + b) = bc(1 - b²) = bc - b³c

Substituting this back into the simplified expression, we get:
(ab - ac)(a + 1) + (bc - b³c) + (bc - c²)



Rearranging the terms, we have:
(ab - ac)(a + 1) + (bc - c²) + (bc - b³c)

Notice that the second term (bc - c²) and the third term (bc -