Question

Find the value of $( a^2 + \sqrt{a^2-1})^4 + ( a^2 - \sqrt{a^2-1})^4$

Answer 1

Firstly, the expression $(x+ y)^ 4+ (x - y)^ 4$ is simplified by using Binomial Theorem.
This can be done as
$(x+y)^4 = \space^4C_0x^4+\space^4C_1x^3y^+ \space^4C_2x^2y^2+\space^4C_3xy^3 +\space^ 4C_4y^4$

$=x^4 + 4x^3y+ 6x^2y^2 + 4xy^3 + y^4$

$(x-y)^4 =\space^ 4C_0x^4 - \space^4C_1x^3y+\space^ 4C_2x^2y^2 - \space^4C_3xy^3 +\space^ 4C_4y^4$

$=x^4 - 4x^3y+ 6x^2y^2 - 4xy^3 + y^4$

$∴ (x+ y)^4 + (x - y)^4 = 2(x^4 + 6 x^2y^2+ y^4)$

Putting $x = a^2$ and $y = \sqrt{a^2-1} =\sqrt{a^2-1}$, we obtain

$(a^2 + \sqrt{a^2 −1})^4 + (a^2 - \sqrt{a^2 −1}) = 2 (a^2)^2 +6(a^2) (\sqrt{a^2−1})^2 + (\sqrt{a^2-1})^4]$

$=2 [a^8 + 6a^4 (a^2 - 1) + (a^2 − 1)^2]$

$= 2[a^8 + 6a^6 − 6a^4 + a^4 − 2a^2+1]$

$=2[a^8 +6a^6 -5a^4 -2a^2+1]$

$= 2a^8 +12a^6 - 10a^4 - 4a^2 +2$