<s>[INST] Radical questions algebra Hello everyone how would I simplify the following radicals.
$$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$$
I got $$3 \sqrt{64a^4b^7}$$ I know $64$ square root is $8$ and $a^4$ square root is $a^2$
My second question is how would I simplify the following
$$\frac{3}{4}\sqrt{3t^3}$$
I know $ \sqrt[\large4]{3t^3}$ is equal to $(3t)^{\frac{3}{4}}$ so would I multiply by $\frac{1}{4}$ 
My final question is how would I simplify the following 
$$ \sqrt[\large 6]{x^6y^4}$$
Whomever helps Fernando with these questions shall receive his eternal gratitude. 
 [/INST] 
$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$

$$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}} = 3 \sqrt{64a^4b^7} = 3\sqrt{64a^4}\sqrt{b^7} = 3\cdot 8 a^2 \sqrt{b^7} = 24a^2 \sqrt{b^7} = 24a^2b^{\large\frac72}\quad\quad\quad\tag{1}$$


"$ \;\sqrt[\large4]{3t^3}\;$ is equal to $\;(3t)^{\frac{3}{4}}\;$"

Not quite:
$$\sqrt[\large 4]{3t^3} \;=\; (3\,t^3)^{\large \frac 14} \;= \;3^{\large \frac{1}{4}}\,t^{\large\frac{3}{4}}\tag{2}$$
EDIT: to address comment/question below
If your original expression (to simplify) was $\;\dfrac{3}{\sqrt[\large 4]{3t^3}}\;$ then using the simplification above, we have $$\;\dfrac{3}{\sqrt[\large 4]{3t^3}},\;= \;\frac{3}{3^{\large \frac{1}{4}}\,t^{\large\frac{3}{4}}}\;=\;\frac{3^{\large\frac{4}{4}}\cdot 3^{-\large\frac{1}{4}}}{t^{\large\frac{3}{4}}} \;=\; \frac{3^{\large\frac{3}{4}}}{t^{\large\frac{3}{4}}}\;=\;\frac{(3^3)^{\large\frac{1}{4}}}{(t^3)^{\large\frac{1}{4}}} \;=\;\left(\frac{27}{t^3}\right)^{\large\frac{1}{4}} \;= \; \sqrt[\large 4]{\frac{27}{t^3}}$$


My final question is how would I simplify the following: $\quad \sqrt[\large 6]{x^6y^4}\;$?

$$\sqrt[\large 6]{x^6y^4}\; = \;\left(x^6y^4\right)^{\large \frac{1}{6}}\; =\; x^{\large \frac{6}{6}}y^{\large \frac{4}{6}} \;=\; xy^{\large \frac{2}{3}}\; = \;x\sqrt[\large 3]{y^2}\tag{3}$$
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