English: From equation your 16 this one we can write this one we can write that $\frac{1}{3}$ (β I_(b) - β² I_(b)) that means, from this is from equation-16 this equation-16, I_(a1) = $\frac{1}{3}$ (1*0 + β I_(b) - β² I_(b)) .
Telugu: సమీకరణం 16 నుండి మనము $\frac{1}{3}$ (β I_(b) - β² I_(b)) అని రాసుకోవచ్చు.