Question

Hi guys gm...
If a+b=5 and a2+b2=11 then prove that a3+b3=20

Answer 1

Identities:

Identity I: (a + b)2 = a2 + 2ab + b2

Identity II: (a – b)2 = a2 – 2ab + b2

Identity III: a2 – b2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab

Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)

Identity VIII: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Solution:

given in the question: a + b = 5, a² + b² = 11

what to prove: a³ + b³ = 20

using the Identity I: (a + b)2 = a2 + 2ab + b2  

5)² = 11 + 2ab

25 = 11 + 2ab

2ab = 25 - 11

2ab = 14

ab = 14 / 2

ab = 7

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

RHS: (a + b)(a² + b² - ab)

Putting given values in it.

We get

= (5)(11 - 7)

= 5 × 4

= 20

= L.H.S.

Hence, proved.

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