The sum of the coefficients of all odd degree terms in the expansion of (x+√(x³))⁵ + (x-√(x³))⁵, (x>1) is

(x+√(x³))⁵+(x-√(x³))⁵
(a+b)⁵=a⁵+5a⁴b+10a³b²+10a²b³+5ab⁴+b⁵
(a-b)⁵=a⁵-5a⁴b+10a³b²-10a²b³+5ab⁴-b⁵
(a+b)⁵+(a-b)⁵=2[a⁵+10a³b²+5ab⁴]
=2[x⁵+10x³(x³)+5x(x³)⁴]
=2[x⁵+10x⁶+5x⁷]
Here all the coefficients of the above equation are odd terms coefficients of even terms are cancelled out. So, sum of coefficients →2+20+10=2