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Question
Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate `(sqrt2 + 1)^6 + (sqrt2 -1)^6`
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Solution
Using Binomial Theorem, the expressions, (x + 1)6 and (x – 1)6, can be expanded as
`(x + 1)^6 = x^6 + ^6C_1 x^5 1^1 + ^6C_2 x^4 xx 1^2 + ^6C_3 x^3 xx 1^3 + ^6C_4 x^2 1^4 + ^6C_5. x. 1^5 + 1^6`
= `x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1`
similarly, `(x -1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1`
on adding `(x - 1)^6 + (x - 1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1)`
Putting x = `sqrt2` in this
`(sqrt2 + 1)^6 + (sqrt2 - 1)^6 = 2[(sqrt2)^6 + 15 (sqrt2)^4 + 15 (sqrt2)^2 + 1]`
= 2[8 + 15 x 4 + 15 x 2 +1]
= 2[8 + 60 + 30 +1]
= 2 x 99
= 198
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