What is t if t^8=8^2 ?

Since we need the formula of difference of squares to
solve the equation, we'll recall first the formula for the difference of
squares:

a^2 - b^2
=(a-b)(a+b)

Now,we'll put a^2 = t^8 = (t^4)^2 and b^2 = 64
= 8^2

We'll re-write the given equation, emphasizing on the
difference of squares:

 (t^4)^2 - 8^2 = (t^4 - 8)(t^4 +
8)

t^4 - 8 is also a difference o squares, whose terms are:
a = t^2 and b = 2sqrt2

(t^4 - 8)(t^4 + 8) = (t^2 -
2sqrt2)(t^2 + 2sqrt2)(t^4 + 8)

Now, we'll solve the
equation:

t^8 - 8^2 = 0

t^8
- 8^2 = (t^2 - 2sqrt2)(t^2 + 2sqrt2)(t^4 + 8)

(t^2 -
2sqrt2)(t^2 + 2sqrt2)(t^4 + 8) = 0

t^2 - 2sqrt2 =
0

t^2 = 2sqrt2

t1 =
+sqrt(2sqrt2)

t2 =
-sqrt(2sqrt2)

t^2 + 2sqrt2 =
0

t3 = +i*sqrt(2sqrt2)

t4 =
-i*sqrt(2sqrt2)

The roots of the equation,
both real and imaginary, are {+sqrt(2sqrt2) ; -sqrt(2sqrt2) ; +i*sqrt(2sqrt2) ;
-i*sqrt(2sqrt2)}.