Question

calculando [a*√a^-1√a^-1√a^-1]sabendo que as raízes estão 1 dentro da outra.

Answer 1

[a*√a^-1  √a^-1  √a^-1] =
[a*√(1/a)^1 √(1/a)^1 √(1/a)^1] = 
[a*√(1/a)^1 √ (1/a)^1 . (1/a)^1/2] = 
[a*√(1/a)^1.√(1/a)^(1 + 1/2)] =
[a*√(1/a)^1.√(1/a)^2/2 + 1/2)] = 
[a*√(1/a)^1.√(1/a)3/2)] = 
[a*√1/a)^1 . (1/a)^3/2/2] =
[a*√(1/a)^1 . (1/a)^3/4] = 
[a*√(1/a)^1 + 3/4] = 
[a*√(1/a^4/4+3/4]=
[a*√(1/a)^7/4] = 
[a*(1/a)^7/4/2] = 
[a*(1/a)^7/4 . 1/2)]=
[a*(1/a)^7/8]=
 [a*(a)^-7/8]=
 a^1 - 7/8 = a^8/8 - 7/8 = a^1/8