@acer 
Thank you very much, it's excellent

@dharr 
Thank you for your comment
Variables a1 have an unknown range, but k and re are positive numbers that are later substituted in the general solution for the variables a1

Thank you for your comment
I know that the number of unknowns in the above system is ten and not eight, but in my studies, k and re are arbitrary constants and not unknowns, so first all solutions of a1 must contain k and re, and then I discuss some special cases of k and re values
As you mentioned in your comment when the values of k and re are determined then fsolve easily gives a solution but that's not what I want to get.

Is there a way to get solutions of a1 with k and re

@Carl Love 

Thank you very much it's great and working properly

@Carl Love 

thank you for replying
Yes, these three assumptions are acceptable

@Carl Love 

I'm trying to doing this rule in which the multiplication of any two functions T(x, a)*T(x, b) must convert  into an addition

1/2*(T(x, a+b)+T(x, abs(a-b)))

on this series for example

p := 8*T(x, 7)*T(x, 2)+4*T(x, 5)*T(x, 1)+6*T(x, 3)*T(x, 3)+7*T(x, 1)*T(x, 4)

After applying this rule T(x,a)*T(x,b) =1/2*(T(x,a+b)+T(x,abs(a-b))) i must get

pp := 4*(T(x, 9)+T(x, 5))+2*(T(x, 6)+T(x, 4))+3*(T(x, 6)+T(x, 0))+7/2*(T(x, 5)+T(x, 3))

Please help. 

@Kitonum 

Thank you very much, i appreciate it

How to convert  this
 T(x,a)*T(x,b) into

( T(x,a+b) +T(x,abs(a-b)) )*1/2

@Kitonum 

   Thank you very much
Can I add 1 for each T(x,i) to become T(x,i+1)

8T(x, 8)+8*T(x, 3)+4*T(x, 6)+11*T(x, 2)+12*T(x, 4)+7*T(x,5) 

or minus one to become

8*T(x, 6)+8*T(x, 1)+4*T(x, 4)+11*T(x, 0)+12*T(x, 2)+7*T(x, 3)