Question

If a/b = b/c and a, b, c > 0, then show that

i. (a + b + c)(b - c) = ab - c²

ii. (a² + b²)(b² + c²) = (ab + be)²

iii. (a² + b²)/ab = (a + c)/b

Answer 1

Let a/b = b/c = k

∴ b = ck 

∴ a = bk = (ck)k 

∴ a = ck² …(ii) 

i. (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)] 

= c²k³ – c² 

= c²(k – 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 

ii. (a + b²)(b + c²) = (ab + bc)²

b = ck; a = ck² 

L.H.S = (a² + b²) (b² + c²) 

= [(ck²) + (ck)²] [(ck)² + c²] … [From (i) and (ii)] 

= [c²k⁴ + c²k²] [c²k² + c²] 

= c²k²(k² + 1) c²(k² + 1) 

= c⁴k²(k² + 1)² 

R.H.S = (ab + bc)²

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

= [c²k³ + c²k²] 

= [c²k(k² + 1)]² = c⁴(k² + 1)² 

∴ L.H.S = R.H.S 

∴ (a² + b²) (b² + c²) = (ab + bc)²