# Find the value of We know that- Hence    = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Thus, (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Replacing b with -b

(a + (-b))4 = a4 + 4a3(-b) + 6a2(-b)2 + 4a(-b)3 + (-b)4

(a-b)4 = a4 - 4a3b + 6a2b2 - 4ab3 + b4

Now,

(a + b)4 + (a - b)4

= (a4 + 4a3b + 6a2b2 + 4ab3 + b4) + (a4 - 4a3b + 6a2b2 - 4ab3 + b4)

= 2(a4 + 6a2b2 + b4)

Putting a = a2 & b = √(a2-1)

(a2 + √(a2-1))4 + (a2 - √(a2-1))4

= 2[(a2)4 + 6(a2)2(√(a2-1))2 + (√(a2-1))4]

= 2[ a8 + 6(a4)(a2-1) + (a2-1)2]

= 2[ a8 + 6 a6 - 6a4 + a4 - 2 a2 + 1]

= 2[ a8 + 6 a6 - 5a4 - 2 a2 + 1]

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