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Question
Find the value of `(a^2 + sqrt(a^2 - 1))^4 + (a^2 - sqrt(a^2 -1))^4`
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Solution
Firstly, the expression (x + y)4 + (x – y)4 is simplified by using Binomial Theorem.
(x + y)4 = 4C0x4 + 4C1x3y + 4C2x2y2 + 4C3xy3 + 4C4y4
= x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x - y)4 = 4C0x4 - 4C1x3y + 4C2 x2y2 - 4C3xy3 + 4C4y4
= x4 - 4x3y + 6x2y2 - 4xy3 + y4
∴ (x + y)4 + (x - y)4 = 2(x4 + 6x2 y2 + y4)
Putting x = a2 and y = `sqrt(a^2 - 1)`, we obtain
`a^2 + sqrt((a^2 - 1)^4) + a^2 - sqrt((a^2 - 1)^4) = 2[(a^2)^4 + 6 (a^2)^2 sqrt((a^2 - 1)^2) + sqrt(a^2 - 1)^4]`
= `2[a^8 + 6a^4 (a^2 - 1) + (a^2 - 1)^2]`
= `2[a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1]`
= `2[a^8 + 6a^6 - 5a^4 - 2a^2 + 1]`
= `2a^8 + 12a^6 - 10a^4 - 4a^2 + 2`
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