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To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!
Next, we will check for extraneous solutions. We do that by substituting $27$ for $x$ into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution.
Because our substitution produced a true statement, we know that our answer, $x=27,$ is correct.

$-3x +3=0$

SubEqn$LHS−3=RHS−3$

$-3x =-3$

MultEqn$LHS⋅(-1)=RHS⋅(-1)$

$3x =3$

RaiseEqn$LHS_{3}=RHS_{3}$

$x=27$