Using the identity (a + b)(a – b) = a2 – b2, find the following p

1.	Using the identity (a + b)(a – b) = a2 – b2, find the following product.(i) (p + 2) (p – 2)(ii) (1 + 3b) (3b – 1)(iii) (4 – mn) (mn + 4)(iv) (6x + 7y) (6x – 7y)
Answer» (i) (p + 2) (p – 2) Substituting a = p; b = 2 in the identity (a + b) (a – b) = a² – b², we get (p + 2) (p – 2) = p² – 2² (ii) (1 + 3b) (3b – 1) (1 + 3b) (3b -1) can be written as (3b + 1) (3b – 1) Substituting a = 36 and b = 1 in the identity (a + b) (a – b) = a² – b², we get (3b + 1) (3b – 1) = (3b)² – 1² = 3² x b² – 1² (3b + 1) (3b – 1) = 9b² – 1² (iii) (4 – mn) (mn + 4) (4 – mn) (mn + 4) can be written as (4 – mn) (4 + mn) = (4 + mn) (4 – mn) Substituting a = 4 and b = mn is (a + b) (a – b) = a² – b², we get (4 + mn) (4 – mn) = 4² – (mn)² = 16 – m² n² (iv) (6x + 7y) (6x – 7y) Substituting a = 6x and b = 7y in (a + b) (a – b) = a² – b², we get (6x + 7y) (6x – 7y) = (6x)² – (7y)² = 6²x² – 7²y² (6x + 7y) (6x – 7y) = (6x)² – (7y)² = 6²x² – 7²y² (6x + 7y) (6x – 7y) = 36x² – 49y²

Using the identity (a + b)(a – b) = a2 – b2, find the following product.(i) (p + 2) (p – 2)(ii) (1 + 3b) (3b – 1)(iii) (4 – mn) (mn + 4)(iv) (6x + 7y) (6x – 7y)

Answer»

(i) (p + 2) (p – 2)

Substituting a = p; b = 2 in the identity

(a + b) (a – b) = a² – b², we get

(p + 2) (p – 2) = p² – 2²

(ii) (1 + 3b) (3b – 1)

(1 + 3b) (3b -1) can be written as (3b + 1) (3b – 1)

Substituting a = 36 and b = 1 in the identity

(a + b) (a – b) = a² – b², we get

(3b + 1) (3b – 1) = (3b)² – 1²

= 3² x b² – 1²

(3b + 1) (3b – 1) = 9b² – 1²

(iii) (4 – mn) (mn + 4)

(4 – mn) (mn + 4) can be written as (4 – mn) (4 + mn) = (4 + mn) (4 – mn)

Substituting a = 4 and b = mn is

(a + b) (a – b) = a² – b², we get

(4 + mn) (4 – mn) = 4² – (mn)²

= 16 – m² n²

(iv) (6x + 7y) (6x – 7y)

Substituting a = 6x and b = 7y in

(a + b) (a – b) = a² – b², we get

(6x + 7y) (6x – 7y) = (6x)² – (7y)²

= 6²x² – 7²y²

(6x + 7y) (6x – 7y) = (6x)² – (7y)²

= 6²x² – 7²y²

(6x + 7y) (6x – 7y) = 36x² – 49y²