#### Answer

The solution is $x=1$

#### Work Step by Step

$x^{3/2}-2x^{3/4}+1=0$
Rewrite $x^{3/2}$ as $(x^{1/2})^{3}$ and $x^{3/4}$ as $(x^{1/4})^{3}$:
$(x^{1/2})^{3}-2(x^{1/4})^{3}+1=0$
Let $u$ be equal to $x^{1/4}$
If $u=x^{1/4}$, then $u^{2}=x^{1/2}$
Rewrite the equation using the new variable $u$:
$(u^{2})^{3}-2u^{3}+1=0$
Simplify:
$u^{6}-2u^{3}+1=0$
Let $z$ be equal to $u^{3}$
If $z=u^{3}$, then $z^{2}=u^{6}$
Rewrite the equation again, using the variable $z$:
$z^{2}-2z+1=0$
Solve by factoring:
$(z-1)^{2}=0$
$\sqrt{(z-1)^{2}}=\sqrt{0}$
$z-1=0$
Solve for $z$:
$z=1$
Substitute $z$ back to $u^{3}$ and solve for $u$:
$u^{3}=1$
$\sqrt[3]{u^{3}}=\sqrt[3]{1}$
$u=1$
Substitute $u$ back to $x^{1/4}$ and solve for $x$:
$x^{1/4}=1$
$(x^{1/4})^{4}=1^{4}$
$x=1$
The solution is $x=1$