A^{2}_{ik} means each member powered to 2 . A_{ij}*A_{jk}=A^{2}_{ik} is true only if A is a diagonal tensor .which I think you didn't notice . thank you so much for your time and effort .
With principal directions in 3d
(s_{11}^{2}+s_{22}^{2}+s_{33}^{2})^{2} = s_{11}^{4}+s_{22}^{4}+s_{33}^{4} + 2(s_{11}^{2}s_{22}^{2}+ s_{11}^{2}s_{33}^{2}+ s_{22}^{2}s_{33}^{2})
(s_{11}+s_{22}+s_{33})^{2} = s_{11}^{2}+s_{22}^{2}+s_{33}^{2} + 2(s_{11}s_{22}+ s_{11}s_{33}+ s_{22}s_{33}) = 0 because s_{ii}=0. So
s_{11}^{2}+s_{22}^{2}+s_{33}^{2} = -2(s_{11}s_{22}+ s_{11}s_{33}+ s_{22}s_{33})
(s_{11}^{2}+s_{22}^{2}+s_{33}^{2})^{2} = 4[s_{11}^{2}s_{22}^{2}+s_{11}^{2}s_{33}^{2}+s_{22}^{2}s_{33}^{2 }+ 2(s_{11}+s_{22}+s_{33})(s_{11}s_{22}s_{33})] and s_{ii}=0. So
s_{11}^{2}s_{22}^{2}+s_{11}^{2}s_{33}^{2}+s_{22}^{2}s_{33}^{2} = ¼(s_{11}^{2}+s_{22}^{2}+s_{33}^{2})^{2} and substitute this into first eq to get:
s_{11}^{4}+s_{22}^{4}+s_{33}^{4 }= ½(s_{11}^{2}+s_{22}^{2}+s_{33}^{2})^{2} , or in indicial notation:
s_{ii}^{4} = s_{ii}^{2}s_{jj}^{2}
The above in indicial notation is:
(s_{ii}^{2})^{2 = }s_{ii}^{4} + 2A
(s_{ii})^{2} = s_{ii}^{2 }+ 2B = 0
s_{ii}^{2 }= -2B
(s_{ii}^{2})^{2} = 4B^{2} = 4A, if you can show B^{2} = A
Then
s_{ii}^{4} = s_{ii}^{2}s_{jj}^{2}
I can’t express things like s_{11}s_{22}+ s_{11}s_{33}+ s_{22}s_{33} in indicial notation.
Surely your Prof at some point has to show you how to do this problem in indicial notation.
Please let us know.
EDIT: Had to juggle with Latex help again to get my Word sub-super script version to post.