In attempting to derive the Mohr circle equations and given $$ \sigma_N=\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2\tag{1}$$ $$\sigma_n^2+\sigma_S^2=\sigma_1^2n_1^2+\sigma_2^2n_2^2+\sigma_3^2n_3^2\tag{2} $$ $$n_1^2+n_2^2+n_3^2=1\tag{3},$$
how does one get from (1), (2) and (3) to (4)? $$ n_1^2 = \frac{(\sigma_N-\sigma_2)(\sigma_N-\sigma_3)+\sigma_S^2}{(\sigma_1-\sigma_2)(\sigma_1-\sigma_3)} $$
Inserting $\sigma_N$ from $(1)$ into $(2)$ gives $$ \sigma_S^2=- ((2n_3^2\sigma_3 - \sigma_1 + 2n_2^2\sigma_2 + n_1^2\sigma_1)n_1^2\sigma_1 + (n_3 + 1)(n_3 - 1)n_3^2\sigma_3^2 + n_2^4\sigma_2^2 + (2n_3^2\sigma_3 - \sigma_2)n_2^2\sigma_2). $$ Then insert $\sigma_N$ and $\sigma_S^2$ into the required equation: $$ \frac{(\sigma_N-\sigma_2)(\sigma_N-\sigma_3)+\sigma_S^2}{(\sigma_1-\sigma_2)(\sigma_1-\sigma_3)}-n_1^2= -\frac{(n_1^2 + n_2^2 + n_3^2 - 1)\sigma_2\sigma_3}{(\sigma_1 - \sigma_2)(\sigma_1 - \sigma_3)} $$ This is obviously equal to zero by equation $(3)$. So we are done.
