(i) (2a – 3b – c)2
We have, (2a – 3b – c)2
Using identity (a + b + c)2
= a2 + b2 + c2 + 2ab + 2bc + 2ca, we get
∴(2a – 3b – c)2
= (2a)2 + (- 3b)2 + (-c)2 + 2(2a)(- 3b) + 2(- 3b)(- c) + 2(- c)2a
= 4a2 + 9b2 + c2 – 12ab + 6bc – 4ac
(ii) (2 + x – 2y)2
We have, (2 + x – 2y)2
= (2)2 + (x)2 + (- 2y)2 + 2(2)(x) + 2(x)(- 2y) + 2(2)(- 2y)
= 4 + x2 + 4y2 + 4x – 4xy – 8y
(iii) (a + 2b + 4c)2
(iv) (m + 2n – 5p)2
(v) (3a – 7b – c)2
(vi) (x/y + y/z + z/x)2
