Given,
Rationalising the above term,
Using the formula (a + b) (a - b) = (a2 – b2)
⸫ 4 + √15
Comparing with a + √15 b,
⸫ a = 4, b = 1
OR
Solution: Given, (5a – 7b)3 + (9c – 5a)3 + (7b – 9c)3
Using the formula, (a + b + c)3 = a3 + b3 + c3 + 3(a + b) (b + c) (c + a)
⇒ a3 + b3 + c3 = (a + b + c)3 - 3(a + b) (b + c) (c + a)
⇒ (5a – 7b)3 + (9c – 5a)3 + (7b – 9c)3 = (5a - 7b + 9c – 5a + 7b – 9c)3 – 3(5a – 7b + 9c – 5a) (9c – 5a + 7b – 9c) (7b - 9c + 5a – 7b)
⇒ (5a – 7b)3 + (9c – 5a)3 + (7b – 9c)3 = 03– 3(-7b + 9c) (-5a + 7b) (-9c + 5a)
⸫ (5a – 7b)3 + (9c – 5a)3 + (7b – 9c)3 = 3(5a – 7b) (7b – 9c) (9c – 5a)