# Prove that (3+2)6+(3-2)6 = 970 - Mathematics and Statistics

Sum

Prove that (sqrt(3) + sqrt(2))^6 + (sqrt(3) - sqrt(2))^6 = 970

#### Solution

(sqrt(3) + sqrt(2))^6 = ""^6"C"_0 (sqrt(3))^6 (sqrt(2))^0 + ""^6"C"_1 (sqrt(3))^5 (sqrt(2))^1 + ""^6"C"_2 (sqrt(3))^4 (sqrt(2))^2 + ""^6"C"_3 (sqrt(3))^3 (sqrt(2))^3 + ""^6"C"_4 (sqrt(3))^2 (sqrt(2))^4 + ""^6"C"_5 (sqrt(3))^1 (sqrt(2))^5 + ""^6"C"_6 (sqrt(3))^0 (sqrt(2))^6

Since, 6C0 = 6C6 = 1, 6C1 = 6C5 = 6, 6C2 = 6C4 = (6 xx 5)/(2 xx 1) = 15, 6C3 = (6 xx 5 xx 4)/(3 xx 2 xx 1) = 20

∴ (sqrt(3) + sqrt(2))^6 = 1(27)(1) + 6(9sqrt(3))(sqrt(2)) + 15(9)(2) + 20(3sqrt(3))(2sqrt(2)) + 15(3)(4) + 6(sqrt(3))(4sqrt(2)) + 1(1)(8)

∴ (sqrt(3) + sqrt(2))^6 = 27 + 54 sqrt(6) + 270 + 120sqrt(6) + 180 + 24sqrt(6) + 8    ...(i)

Also, (sqrt(3) - sqrt(2))^6 = ""^6"C"_0 (sqrt(3))^6 (sqrt(2))^0 - ""^6"C"_1 (sqrt(3))^5 (sqrt(2))^1 + ""^6"C"_2 (sqrt(3))^4 (sqrt(2))^2 - ""^6"C"_3 (sqrt(3))^3 (sqrt(2))^3 + ""^6"C"_4 (sqrt(3))^2 (sqrt(2))^4 - ""^6"C"_5 (sqrt(3))^1 (sqrt(2))^5 + ""^6"C"_6 (sqrt(3))^0 (sqrt(2))^6

= 1(27)(1) - 6(9sqrt(3)) (sqrt(2)) + 15(9)(2) - 20(3sqrt(3)) (2sqrt(2)) + 15(3)(4) - 6(sqrt(3)) (4sqrt(2)) + 1(1)(8)

∴ (sqrt(3) - sqrt(2))^6 = 27 - 54sqrt(6) + 270 - 120sqrt(6) + 180 - 24sqrt(6) + 8   ...(ii)

Adding (i) and (ii), we get

(sqrt(3) + sqrt(2))^6 + (sqrt(3) - sqrt(2))^6 = (27 + 54sqrt(6) + 270 + 120sqrt(6) + 180 + 24sqrt(6) + 8) + (27 - 54sqrt(6) + 270 - 120sqrt(6) + 180 - 24sqrt(6) + 8)

= 54 + 540 + 360 + 16

= 970

Concept: Binomial Theorem for Positive Integral Index
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.2 | Q 4. (i) | Page 77