Fiven the matrix `A= [[-1,3,5],[1,-3,-5],[-1,3,5]]` and X be the solution set of the equation `A^(x)=A,` where `x in N -{1}`. Evaluate `prod((x^(3)+1)/(x^(3)-1))` where the contincued extends for all `x in X`.

Explanation

`A= [[-1,3,5],[1,-3,-5],[-1,3,5]][[-1,3,5],[1,-3,-5],[-1,3,5]]=[[-1,3,5],[1,-3,-5],[-1,3,5]]`
= A
`therefore A^(2) = A^(3) = A^(4) = A^(5) = …= A`
but given `A^(x) = A`
`rArr x = 2, ,3, 4, 5, … [because x ne 1,` given]
`therefore prod ((x^(3)+1)/(x^(3)-1)) = prod ((x+1)/(x-1)) prod (x^(2)-x+1)/(x^(2)+x+1))` On putting `x = 2, 3, 4, 5...`
`prod ((x^(3)+1)/(x^(3)-1)) =lim_(nrarr infty) prod_(x=2)^(n) ((x+1)/(x-1)) prod_(x=2)^(n) (x^(2)-x+1)/(x^(2)+x+1))`
`lim _(n rarr infty)((3cdot 4cdot 5...(n-1) n (n+1) )/(1cdot 2cdot 3 ... (n-3) (n-2)(n-1)))`
`xxlim _(n rarr infty)((3cdot7cdot... (n^(2)-m+1) )/(7cdot 13 ... (n^(2)-n+1) (n^(2) + n+1)))`
`=lim _(n rarr infty) (n(n+1))/(2)xx 3/(n^(2)+n+1)`
`= 3/2lim _(n rarr infty) ((1+1/n))/((1+1/n+1/n^(2)))= 3/2cdot ((1+0))/((1+0+0))=3/2`