The sum of the coefficients of all even degree terms is x in the expansion of x+x3−16+x−x3−16 (x>1)  is equa to

Given expression is x+x3−16+x−x3−16=2 6C0x6+6C2x4x3−12+6C4x2x3−14+6C6x3−16∵(a+b)n+(a−b)n=2 nC0an+nC2an−2b2+nC4an−4b4+… 6C0x6+6C2x4x3−1+6C4x2x3−12+6C6x3−13 The sum of the terms with even power of x=2 6C0x6+6C2−x4+6C4x8+6C4x2+6C6−1−3x6 =2 6C0x6−6C2x4+6C4x8+6C4x2−1−3x6Now, the required sum of the coefficients of even powers of x in x+x3-16+x−x3-16=2 6C0−6C2+6C4+6C4−1−3=2[1−15+15+15−1−3]=2(15−3)=24