#### Answer

$x=\left\{ -27,3 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
(x^2+24x)^{1/4}=3
,$ raise both sides to the exponent equal to $ 4 .$ Then express the resulting equation in the form $ax^2+bx+c=0,$ and use the concepts of factoring quadratic equations to solve the resulting equation. Finally, do checking if the solution satisfies the original equation.
$\bf{\text{Solution Details:}}$
Raising both sides to the fourth power, the given equation becomes
\begin{array}{l}\require{cancel} x^2+24x=81 \\\\ x^2+24x-81=0 .\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 1(-81)=-81 $ and the value of $b$ is $ 24 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -3,27 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel}
x^2-3x+27x-81=0 .\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel}
(x^2-3x)+(27x-81)=0
.\end{array}
Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel}
x(x-3)+27(x-3)=0 .\end{array}
Factoring the $GCF= (x-3) $ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x-3)(x+27)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-3=0 \\\\\text{OR}\\\\ x+27=0 .\end{array}
Solving each equation results to \begin{array}{l}\require{cancel}
x-3=0 \\\\ x=3 \\\\\text{OR}\\\\ x+27=0 \\\\ x=-27 .\end{array}
Upon checking, $
x=\left\{ -27,3 \right\}
$ satisfy the original equation.