Solution
= Solution
In three dimensions each contracted <Kronecker delta> has value three, while $b_kb_k=|\mathbf b|^2$. Thus
$$
\boxed{\delta_{jj}b_kb_k+\delta_{ii}=3|\mathbf b|^2+3}.
$$
= Solution
In three dimensions each contracted <Kronecker delta> has value three, while $b_kb_k=|\mathbf b|^2$. Thus
$$
\boxed{\delta_{jj}b_kb_k+\delta_{ii}=3|\mathbf b|^2+3}.
$$