Question

If a/b = b/c and a, b, c > 0, then show that  (a + b + c)(b – c) = ab –c²

Answer 1

Let a/b = b/c =  k 

∴ b = ck 

∴ a = bk =(ck)k 

∴ a = ck …(ii)

(a + b + c)(b – c) = ab – c² 

L.H.S = (a + b + c) (b – c) 

= [ck² + ck + c] [ck – c] … [From (i) and (ii)] 

= c(k² + k + 1) c (k – 1) 

= c² (k² + k + 1) (k – 1) 

R.H.S = ab – c² 

= (ck² ) (ck) – c² … [From (i) and (ii)] 

= c² k³ – c² 

= c² (k³ – 1) 

= c² (k – 1) (k² + k + 1) … [a³ – b³ = (a – b) (a² + ab + b² ] 

∴ L.H.S = R.H.S

∴ (a + b + c) (b – c) = ab – c²