Question

if a/b=b/c and a, b, c > 0 then show that a^2+b^2/ab=a+c/b​

Answer 1

Step-by-step explanation:

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

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

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