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Misc 5 - Evaluate (root 3 + root 2)^6 - Chapter 8 Class 11 Binomial

Misc 5 - Chapter 8 Class 11 Binomial Theorem - Part 2
Misc 5 - Chapter 8 Class 11 Binomial Theorem - Part 3
Misc 5 - Chapter 8 Class 11 Binomial Theorem - Part 4
Misc 5 - Chapter 8 Class 11 Binomial Theorem - Part 5

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Misc 5 Evaluate (√3 + √2 )6 – (√3 – √2 )6 . Finding (a + b)6 – (a – b)6 We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence (a + b)6 = = 6!/(0! (6 − 0)!) a6 × 1 + 6!/(1! (6 − 1) !) a5 b + 6!/2!(6 − 2)! a4 b2 + 6!/3!(6 − 3)! a3 b3 + 6!/(4! (6 − 4) !)a2 b4 + 6!/5!(6 − 5)! ab5 + 6!/(6 ! (6 − 6) !) b6 = 6!/(1 × 6! ) a6 + 6!/(1 × 5!) a5 b + 6!/(2! × 4!) a4 b2 + 6!/(3! 3!) a3 b3 + 6!/(4! 2!) a2 b4 + 6!/(5! × 1) a b5 + 6!/(6! × 1) b6 = 6!/6! a6 + (6 ×5!)/(5! ) a5b + (6 × 5 × 4!)/(2 × 4!) a4 b2 + (6 × 5 × 4 × 3!)/(3 × 2 × 1 × 3!) a3 b3 + (6 × 5 × 4!)/(2 × 1 × 4!) a2 b4 + (6 × 5!)/(1 × 5!) ab5 + 6!/6! b6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 For (a – b)6 Replace b by (–b) in (1) (a + (–b)) 6 = a6 + 6a5 (–b) + 15a4 (– b)2 + 20a3 (– b)3 + 15a2 (– b)4 + 6a(– b)5 + (– b)6 (a – b)6 = a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6 Now, (a + b)6 − (a – b)6 = (a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6) − (a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6) = a6 − a6 + 15a4 b2 − 15a4 b2 + 15a2 b4 − 15a2 b4 + b6 − b6 + 6a5 b + 6a5b + 20a3 b3 + 20a3 b3 + 6ab5 + 6ab5 = 12a5 b + 40a3 b3 + 12ab5 = 4ab (3a4 + 10a2b2 + 3b4) Thus, (a + b)6 – (a – b)6 = 4ab (3a4 + 10a2b2 + 3b4) We need to find (√3 + √2)6 – (√3 – √2)6 Putting a = √3 and b =√2 (√3 + √2)6 – (√3 – √2)6 = 4(√3 )(√2 ) (3(√3 )^4+10(√3 )^2 (√2 )^2+3(√2 )^4 ) = 4(√(3×2) ) (3(√3 )^4+10(√3 )^2 (√2 )^2+3(√2 )^4 ) = 4(√6 ) (3(√3 )^4+10(√3 )^2 (√2 )^2+3(√2 )^4 ) = 4(√6 ) (3(3)^(1/2 × 4)+10(3)^(1/2 × 2) (2)^(1/2 × 2)+3(2)^(1/2 × 4) ) = 4(√6 ) (3(3)^2+10(3)^1 (2)^1+3(2)^2 ) = 4(√6 ) (27+60+12) = 4(√6 ) (99) = 396√6 Thus, (√3 + √2)6 – (√3 – √2)6 = 3𝟗𝟔√𝟔

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.