Answer
$\left\{-\dfrac{64}{27},-1,1,\dfrac{64}{27}\right\}$
Work Step by Step
Let $z=
t^{2/3}
$. Then the given equation, $
9t^{4/3}-25t^{2/3}+16=0
,$ is equivalent to
\begin{align*}
9\left(t^{2/3}\right)^2-25t^{2/3}+16&=0
\\
9z^2-25z+16&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-1)(9z-16)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-1=0 & 9z-16=0
\\
z=1 & 9z=16
\\\\
& z=\dfrac{16}{9}
.\end{array}
Since $z=
t^{2/3}
,$ by back substitution, then
\begin{array}{l|r}
t^{2/3}=1 & t^{2/3}=\dfrac{16}{9}
\\\\
\left(t^{\frac{2}{3}}\right)^3=(1)^3 & \left(t^{\frac{2}{3}}\right)^3=\left(\dfrac{16}{9}\right)^3
\\\\
t^2=1 & t^2=\dfrac{4096}{729}
\\\\
t=\pm\sqrt{1} & t=\pm\sqrt{\dfrac{4096}{729}}
\\\\
t=\pm1 & t=\pm\dfrac{64}{27}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }t=\pm1: & \text{If }t=\pm\dfrac{64}{27}
\\\\
9(\pm1)^{4/3}-25(\pm1)^{2/3}+16\overset{?}=0 &
9\left(\pm\dfrac{64}{27}\right)^{4/3}-25\left(\pm\dfrac{64}{27}\right)^{2/3}+16\overset{?}=0
\\\\
9\left(\sqrt[3]{\pm1}\right)^4-25\left(\sqrt[3]{\pm1}\right)^2+16\overset{?}=0 &
9\left(\sqrt[3]{\pm\dfrac{64}{27}}\right)^4-25\left(\sqrt[3]{\pm\dfrac{64}{27}}\right)^2+16\overset{?}=0
\\\\
9\left(\pm1\right)^4-25\left(\pm1\right)^2+16\overset{?}=0 &
9\left(\pm\dfrac{4}{3}\right)^4-25\left(\pm\dfrac{4}{3}\right)^2+16\overset{?}=0
\\\\
9\left(1\right)-25\left(1\right)+16\overset{?}=0 &
9\left(\dfrac{256}{81}\right)-25\left(\dfrac{16}{9}\right)+16\overset{?}=0
\\\\
9-25+16\overset{?}=0 &
\dfrac{256}{9}-\dfrac{400}{9}+16\overset{?}=0
\\\\
0\overset{\checkmark}=0 &
-\dfrac{144}{9}+16\overset{?}=0
\\\\
&
-16+16\overset{?}=0
\\
&
0\overset{\checkmark}=0
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
9t^{4/3}-25t^{2/3}+16=0
$ is $\left\{-\dfrac{64}{27},-1,1,\dfrac{64}{27}\right\}$.
