Find the value of `(a^2+sqrt(a^2-1))^4+(a^2-sqrt(a^2-1))^4`.

Explanation

We know that

`(a+b)^n=^nC_0a^n+^nC_1a^(n-1)b^1+^nC_2a^(n-2)b^2+.......+^nC_(n-1)a^1b^(n-1)+^nC_nb^n`

Hence,
`(a+b)^4=^4C_0a^4+^4C_1a^3b^1+^4C_2a^2b^2+^4C_3a^1b^3+^4C_4b^4`

`(a+b)^4=(4!)/(0!(4-0)!)a^4+(4!)/(1!(4-1)!)a^3b^1+(4!)/(2!(4-2)!)a^2b^2+(4!)/(3!(4-3)!)a^1b^3+(4!)/(4!(4-4)!)b^4`

`(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4`

Replacing `b` with `-b`

`(a+(-b))^4=a^4+4a^3(-b)+6a^2(-b)^2+4a(-b)^3+(-b)^4`

`(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4`

We need to find `(a+b)^4+(a-b)^4`

`(a+b)^4+(a-b)^4=(a^4+4a^3b+6a^2b^2+4ab^3+b^4)+(a^4-4a^3b+6a^2b^2-4ab^3+b^4)`

`(a+b)^4+(a-b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4+a^4-4a^3b+6a^2b^2-4ab^3+b^4`

`(a+b)^4+(a-b)^4=2a^4+12a^2b^2+2b^4`

`(a+b)^4+(a-b)^4=2(a^4+6a^2b^2+b^4)`

We need to find `(a^2+sqrt(a^2-1))^2+(a^2-sqrt(a^2-1))^2`

`(a^2+sqrt(a^2-1))^2+(a^2-sqrt(a^2-1))^2`

` " " " " = 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)^(1/2xx2)+(a^2-1)^(1/2xx4))`

` " " " " = 2(a^8+6a^4(a^2-1)^(1)+(a^2-1)^2)`

` " " " " =2a^8+12a^4(a^2-1)+2(a^4-2a^2+1)`

` " " " " =2a^8+12a^6-12a^4+2a^4-4a^2+2`

` " " " " =2a^8+12a^6-10a^4-4a^2+2`

So,`(a^2+sqrt(a^2-1))^2+(a^2-sqrt(a^2-1))^2 =2a^8+12a^6-10a^4-4a^2+2`