0
8
6
4
Given (x^33 + 3x)^8
General term = T_{r+1} = 8C_r (x^33)^{8-r} (3x)^r
Since n is even then the middle term is when r = n/2
Here, n/2 = 8/2 = 4
T_5 = 8C_4 (x^33)^4 (3x)^4 = 70 x^{132} 81x^4 = 5670
=> 70 x^{136} = 5670
=> x^{136} = 81
=> x^4 = 3
=> x = ±∛3
Hence, The sum of the real values of x are 0