Answer
$-\sqrt[3]{t}+5\sqrt{t}$
Work Step by Step
The expression $\sqrt[3]{8t}-\sqrt[3]{27t}+\sqrt{25t}$ can be written as,
$\sqrt[3]{8t}-\sqrt[3]{27t}+\sqrt{25t}=\sqrt[3]{{{\left( 2 \right)}^{3}}t}-\sqrt[3]{{{\left( 3 \right)}^{3}}t}+\sqrt{{{\left( 5 \right)}^{2}}t}$
Now, apply the radical of a product property $\sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}$ in the expression:
$\sqrt[3]{{{\left( 2 \right)}^{3}}t}-\sqrt[3]{{{\left( 3 \right)}^{3}}t}+\sqrt{{{\left( 5 \right)}^{2}}t}$
Thus,
$\begin{align}
& \sqrt[3]{8t}-\sqrt[3]{27t}+\sqrt{25t}=\sqrt[3]{{{\left( 2 \right)}^{3}}t}-\sqrt[3]{{{\left( 3 \right)}^{3}}t}+\sqrt{{{\left( 5 \right)}^{2}}t} \\
& =\sqrt[3]{{{\left( 2 \right)}^{3}}}\cdot \sqrt[3]{t}-\sqrt[3]{{{\left( 3 \right)}^{3}}}\cdot \sqrt[3]{t}+\sqrt{{{\left( 5 \right)}^{2}}}\cdot \sqrt{t} \\
&
\end{align}$
Now, apply the power of a radical property $\sqrt[n]{{{a}^{n}}}=a$ $\sqrt{{{\left( 5 \right)}^{2}}},\sqrt[3]{{{\left( 3 \right)}^{3}}},\sqrt[3]{{{\left( 2 \right)}^{3}}}$
Thus,
$\begin{align}
& \sqrt{{{\left( 5 \right)}^{2}}}=5 \\
& \text{ }\sqrt[3]{{{\left( 3 \right)}^{3}}}=3\text{ } \\
& \sqrt[3]{{{\left( 2 \right)}^{3}}}=2
\end{align}$
Substitute the value of$\sqrt{{{\left( 5 \right)}^{2}}},\sqrt[3]{{{\left( 3 \right)}^{3}}},\sqrt[3]{{{\left( 2 \right)}^{3}}}$ in the expression:
$\sqrt[3]{8t}-\sqrt[3]{27t}+\sqrt{25t}=\sqrt[3]{{{\left( 2 \right)}^{3}}}\cdot \sqrt[3]{t}-\sqrt[3]{{{\left( 3 \right)}^{3}}}\cdot \sqrt[3]{t}+\sqrt{{{\left( 5 \right)}^{2}}}\cdot \sqrt{t}$
Thus,
$\begin{align}
& \sqrt[3]{8t}-\sqrt[3]{27t}+\sqrt{25t}=\sqrt[3]{{{\left( 2 \right)}^{3}}}\cdot \sqrt[3]{t}-\sqrt[3]{{{\left( 3 \right)}^{3}}}\cdot \sqrt[3]{t}+\sqrt{{{\left( 5 \right)}^{2}}}\cdot \sqrt{t} \\
& =2\sqrt[3]{t}-3\sqrt[3]{t}+5\sqrt{t} \\
& =-\sqrt[3]{t}+5\sqrt{t}
\end{align}$
Therefore, the solution of the expression $\sqrt[3]{8t}-\sqrt[3]{27t}+\sqrt{25t}$ is$-\sqrt[3]{t}+5\sqrt{t}$.
