Given information: (a-b), (b-c), (c-a) are in G.P.

To find: the value of (a b c)2-3(ab bc ca)

Solution:

Let the common ratio of the G.P. be r.

Then, we have:

(b-c)/(a-b) = (c-a)/(b-c) = r

Simplifying the first equation, we get:

(b-c)² = r(a-b)(c-a)

Expanding the terms, we get:

b² - 2bc + c² = rac - rbc - abr + ar²

Simplifying and rearranging the terms, we get:

r²a + (c-b)r + b² - ac = 0

Using the quadratic formula, we get:

r = [(b-c) ± √((c-b)² - 4(a-b)(c-a))]/2(a-b)

Simplifying the discriminant, we get:

(c-b)² - 4(a-b)(c-a) = 4b² - 4ac

Substituting the value of r in terms of a, b, and c, we get:

r = [(b-c) ± √(4b² - 4ac)]/2(a-b)

Simplifying further, we get:

r = [(b-c) ± 2√(b² - ac)]/2(a-b)

We can now use this expression for r to simplify the given expression:

(a b c)² - 3(ab bc ca)

= a²b² + b²c² + c²a² - 3abc(a + b + c)

= a²b² + b²c² + c²a² - 3abc(a-b + b-c + c-a)

= a²b² + b²c² + c²a² - 3abc(r(a-b) + r(b-c) + r(c-a))

= a²b² + b²c² + c²a² - 3abc[(b-c)² + r(b-c)(c-a)]

Substituting the expression for r, we get:

= a²b² + b²c² + c²a² - 3abc[(b-c)² + (b-c)√(b² - ac)]/2(a-b)(c-a)

= a²b² + b²c² + c²a² - 3abc(b-c)²/2(a-b)(c-a) - 3abc(b-c)√(b² - ac)/2(a-b)(c-a)

Using the identity (a-b)(c-a) = (b-c)² + (b-c)(c-a), we can simplify the expression further:

= a²b² + b²c² + c²a² - 3abc[(b-c)² + (b-c)(c-a)]/2(b-c)² - 3abc(b-c)√(b² - ac)/2(b-c)²

= a²b² + b²c² + c²a² - 3abc - 3abc√(b² - ac)/(b-c)

Now, using the expression for r, we can show that:

r² = (b-c)²/(a-b)(c-a