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This solution is from Arun Iyer, SIA High School and Junior College.
The given equation is $x^{3/2} - 8x^{-3/2} = 7$.
Multiply throughout by $x^{3/2}$and rearranging we get
$(x^{3/2})^2 - 7x^{3/2} - 8 = 0$
This can be factorised as:
$(x^{3/2} - 8)(x^{3/2} + 1) = 0$
Case 1:
Consider,
$x^{3/2} + 1 = 0$ or $x^{3/2} = -1$
Squaring both sides, $x^3 = 1$ so the three cube roots of unity are solutions.
Now
$ 1 = e^{2k\pi i} $
(for $k=0,1,2,3$...) therefore $ x = e^{{2k\pi i} /3} $ .Putting $k=0$, $1$, $2$ we get:
$x = e^0$ (which is $1$) or
$ x = exp({\pm({2\pi i} /3)}) $
.Case 2:
Consider,
$x^{3/2}- 8 = 0$ or $x^{3/2} = 8$
Squaring both sides, $x^3 = 64$ so the three cube roots of $64$ are solutions.
Now
$ x^3 = 64 e^{2k\pi i} $
Putting $k=0,1,2$ we get: $x = 4$ or
$ x = 4exp({\pm{2\pi i} /3}) $
.So the six roots are:
Substituting $x=1$ into the given equation we need to recognise that $1^{3/2}$ has two values $+1$ and $-1$ so that whereas $x^{3/2} = 1$ does not satisfy the equation the other value $x^{3/2} = -1$ does satisfy it and hence $x=1$ is a solution of the equation.