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Mathematics Question on Binomial theorem

_____________If α\alpha and β\beta be the coefficients of x4x^4 and x2x^2 respectively in the expansion of (x+x2−1)6+(x−x2−1)6,\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}, then :

A

α+β=−30\alpha+\beta=-30

B

α−β=−132\alpha-\beta=-132

C

α+β=60\alpha+\beta=60

D

α−β=60\alpha-\beta=60

Answer

α−β=−132\alpha-\beta=-132

Explanation

Solution

The correct answer is B:α−β=−132\alpha-\beta=-132
Given that;
α,β\alpha,\beta are coefficient of x4and  x2x^4 and\space x^2,and;
x+(x2−1)6+x−(x2−1)2x+\sqrt{(x^2-1)^6}+x-\sqrt{(x^2-1)^2}
2[6C0.x6+6C2x4(x2−1)+6C4x2(x2−1)2+6C6(x2−1)3]2\left[^{6}C_{0}.x^{6}+^{6}C_{2}x^{4}\left(x^{2}-1\right)+^{6}C_{4}x^{2}\left(x^{2}-1\right)^{2}+^{6}C_{6}\left(x^{2}-1\right)^{3}\right]
=2[x6+15(x6−x4)+15x2(x4−2x2+1)+(−1+3x2−3x2+x6)]=2[x^6+15(x^6-x^4)+15x^2(x^4-2x^2+1)+(-1+3x^2-3x^2+x^6)]
=64x6−96x4+36x2−2=64x^6-96x^4+36x^2-2
as α,β\alpha,\beta are the coefficient of x4,x2x^4,x^2
∴α=−96 & β=36\therefore \alpha = -96\,\&\,\beta = 36
∴α−β=−132\therefore \alpha - \beta = -132
coefficient