91Ó°ÊÓ

We have given,

$$\begin{gathered} \left(a\right)\sqrt{48}+\sqrt{27} \\ \left(b\right)\sqrt[3]{54}+\sqrt[3]{128} \\ \left(c\right)6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{320} \end{gathered}$$

Here we are going to factor the number inside the radical such that at least on of them is perfect square.

Also,

$\sqrt[n]{a^n}=a$

We have,

$\sqrt{48}+\sqrt{27}$

We will factor 48 and 27,

$48=4^2·3and27=3^2·3$

Thus,

$\sqrt{48}+\sqrt{27}=\sqrt{4^2}·\sqrt{3}+\sqrt{3^2}·\sqrt{3}$

Factoring common factor out,

$\sqrt{48}+\sqrt{27}=(4+3)\sqrt{3}$

$⇒\sqrt{48}+\sqrt{27}=7\sqrt{3}$

Hence, value of, $\sqrt{48}+\sqrt{27}$is $7\sqrt{3}$.

We have,

$\sqrt[3]{54}+\sqrt[3]{128}$

We will factor 54 and 128 such that one of them factor is perfect cube.

$$\begin{gathered} 54=27×2=3^3×2 \\ 128=64×2=4^3×2 \end{gathered}$$

Thus,

$\sqrt[3]{54}+\sqrt[3]{128}=\sqrt[3]{3^3}·\sqrt[3]{2}+\sqrt[3]{4^3}·\sqrt[3]{2}$

$⇒\sqrt[3]{54}+\sqrt[3]{128}=(3+4)\sqrt[3]{2}$

$⇒\sqrt[3]{54}+\sqrt[3]{128}=7\sqrt[3]{2}$

Hence, value of$\sqrt[3]{54}+\sqrt[3]{128}$is$7\sqrt[3]{2}$.

We have,

$6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{320}$

We will factor 320,

role="math" localid="1645083860506" $320=64×5=2^4×20$

$6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{320}=6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{2^4}·\sqrt[4]{20}$

$⇒6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{320}=6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{2^4}·\sqrt[4]{5}·\sqrt[4]{4}$

$⇒6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{320}=\sqrt[4]{5}(3-\sqrt[4]{4})$

Hence, value of$6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{320}$is$\sqrt[4]{5}(3-\sqrt[4]{4})$.