(i) (3x + 7)2 - 84x = (3x - 7)2
L.H.S = (3x + 7)2 – 84x
= (3x)2 + (7)2 + 2 (3x) (7) – 84x
= (3x)2 + (7)2 + 42x – 84x
= (3x)2 + (7)2 – 42x
= (3x)2 + (7)2 – 2 (3x) (7)
= (3x – 7)2
= R.H.S
Hence, proved
(ii) (9a - 5b)2 + 180ab = (9a + 5b)2
L.H.S = (9a – 5b)2 + 180ab
= (9a)2 + (5b)2 – 2 (9a) (5b) + 180ab
= (9a)2 6 (5b)2 – 90ab + 180ab
= (9a)2 + (5b)2 + 9ab
= (9a)2 + (5b)2 + 2 (9a) (5b)
= (9a + 5b)2
= R.H.S
Hence, proved
(iii) \((\frac{4m}{3}-\frac{3n}{4})^2+2mn = \frac{16m^2}{9}+\frac{9n^2}{16}\)
L.H.S = (\(\frac{4m}{3}\) - \(\frac{3n}{4}\))2 + 2mn
= (\(\frac{4m}{3}\))2 + (\(\frac{3n}{4}\))2 – 2mn + 2mn
= (\(\frac{4m}{3}\))2 + (\(\frac{3n}{4}\))2
= \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2
= R.H.S
Hence, verified
(iv) (4pq + 3q)2 - (4pq - 3q)2 = 48pq2
L.H.S = (4pq + 3q)2 – (4pq – 3q)2
= (4pq)2 + (3q)2 + 2 (4pq) (3q) – (4pq)2 – (3q)2 + 24pq2
= 24pq2 + 24pq2
= 48pq2
Hence, proved
(v) (a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) = 0
L.H.S = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
Using identity:
(a – b) (a + b) = a2 – b2
We get,
= (a2 – b2) + (b2 – c2) + (c2 – a2)
= a2 – b2 + b2 – c2 + c2 – a2
= 0
= R.H.S
Hence, verified
