Q. 3 I3.6( 8 Votes )

# Let’s find G.C.D of the following algebraic expression:a2-b2-c2 + 2bc, b2-c2-a2 + 2ac, c2- a2-b2 + 2ab

Factors of a2-b2-c2 + 2bc

= a2-(b2 + c2-2bc)

= a2-(b-c)2

=(a + b-c)(a-b + c) … (Using a2-b2=(a + b)(a-b))

Factors of b2-c2-a2 + 2ac

= b2-(c2 + a2-2ac)

=b2-(c-a)2

=(b + c-a)(b-c + a) … (Using a2-b2=(a + b)(a-b))

Factors of c2-a2-b2 + 2ab

= c2-(a2 + b2-2ab)

=c2-(a-b)2

=(c + a-b)(c-a + b) … (Using a2-b2=(a + b)(a-b))

GCD is the greatest common divisor, which is equal to the product of all the common divisors.

the GCD of a2-b2-c2 + 2bc, b2-c2-a2 + 2ac, c2- a2-b2 + 2ab is 1 as no other factor is common to all the three terms

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