# Misc 5 - Class 11

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 5 Evaluate (3 + 2 )6 – (3 – 2 )6 . First we find (a + b)6 – (a – b)6 and then solve We know that (a + b)n = nC0 an b0 + nC1 an – 1 b1 + nC2 an – 2 b2 + …. …. + nCn – 1 a1 bn – 1 + nCn a0 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 ! 1 × 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 + (3 × 5)a4b2 + (5 × 4)a3b3 + (3 × 5)a2b4 + 6ab5 + b6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 Thus, (a + b)6 = 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 + 6a5 b + 6a5b + 15a4 b2 – 15a4 b2 + 20a3 b3 + 20a3 b3 + 15a2 b4 – 15a2 b4 + 6ab5 + 6ab5 + b6 – b6 = 0 + 12a5b + 0 + 40a3b3 + 0 + 12ab5 + 0 = 12a5b + 40a3b3 + 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 = 43 2 33 4+103 22 2+32 4 = 43×2 33 4+103 22 2+32 4 = 46 33 4+103 22 2+32 4 = 46 3312 × 4+10312 × 2212 × 2+3212 × 4 = 46 332+103121+322 = 46 27+60+12 = 46 99 = 3966 Thus, (3 + 2)6 – (3 – 2)6 = 3966

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.