Show that cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an A.P.
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Sol. If possible let p 1/3 , q 1/3 , r 1/3 be three terms of an A.P. where p, q, r be three distinct prime numbers. Then p 1/3 = a, q 1/3 = a + md, r 1/3 = a + nd; where m, n must be positive integers. On subtracting these equations pair-wise and dividing, we get
= 
⇒ mr 1/3 –nq 1/3 =(m–n) p 1/3 ⇒ (mr
1/3 –nq 1/3 ) 3 = (m–n) 3 p
⇒ m 3 r –n 3 q –3mnr 1/3 q 1/3 (m–n) p 1/3 = (m–n) 3 p
⇒ p 1/3 q 1/3 r 1/3 = 
⇒ irrational = rational
( p, q, r are distinct prime, p 1/3 q 1/3 r 1/3 can not become rational). Which is a contradiction.
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