[tex]\bf (8x-8)^{\frac{3}{2}}=64~~
\begin{cases}
64=2\cdot 2\cdot 2\\
\qquad 2^3
\end{cases}\implies (8x-8)^{\frac{3}{2}}=2^3
\\\\\\
\textit{then we raise both sides by }\frac{2}{3}\implies \left[ (8x-8)^{\frac{3}{2}} \right]^{\frac{2}{3}}=(2^3)^{\frac{2}{3}}
\\\\\\
(8x-8)^1=2^2\implies 8x-8=4\implies 8x=12
\\\\\\
x=\cfrac{12}{8}\implies x=\cfrac{3}{2}[/tex]
now onto the second one,
[tex]\bf (45-3x)^{\frac{1}{2}}=x-9\impliedby \textit{we'll raise both by }^2
\\\\\\
\left[ (45-3x)^{\frac{1}{2}} \right]^2=(x-9)^2\implies (45-3x)^1=x^2-18x+81
\\\\\\
45-3x=x^2-18x+81\implies 0=x^2-15x+36
\\\\\\
0=(x-12)(x-3)\implies x=
\begin{cases}
12\\
3
\end{cases}[/tex]
now, the extraneous part, I don't see either as extraneous, 3 and 12 do work, however, I take it is referring to the root, so if we nevermind the root has ± versions, and say only assuming the value coming from the root is say positive, then
[tex]\bf (45-3x)^{\frac{1}{2}}=x-9\implies \sqrt{45-3x}=x-9\implies \stackrel{x=3}{\sqrt{45-3(3)}=3-9}
\\\\\\
\sqrt{36}=-6\implies \stackrel{\textit{extraneous}}{6\ne -6}[/tex]
