Radical questions algebra

Question

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.

Answer 1

Hint:$$\sqrt[6]{x^6y^4} = \left(x^6y^4\right)^{\frac{1}{6}} = x^\frac{6}{6}y^\frac{4}{6}$$

Answer 2

$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$

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

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"$ \;\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 $$\;\dfrac{3}{\sqrt[\large 4]{3t^3}},\;= \;\frac{3}{3^{\large \frac{1}{4}}\,t^{\large\frac{3}{4}}}\;=\;\frac{3^{\large\frac{4}{4}}\cdot 3^{-\large\frac{1}{4}}}{t^{\large\frac{3}{4}}} \;=\; \frac{3^{\large\frac{3}{4}}}{t^{\large\frac{3}{4}}}\;=\;\frac{(3^3)^{\large\frac{1}{4}}}{(t^3)^{\large\frac{1}{4}}} \;=\;\left(\frac{27}{t^3}\right)^{\large\frac{1}{4}} \;= \; \sqrt[\large 4]{\frac{27}{t^3}}$$

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

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

Answer 3

$3\sqrt{2a^{3}b^{5}}\sqrt{32ab^{2}} = 3(2a^{3}b^{5})^{1/2}(2^{5}ab^{2})^{1/2} = 3(2^{6}a^{4}b^{7})^{1/2} = 3 \cdot 2^{3}a^{2}b^{7/2}$.

Answer 4

To get to the heart of the matter, since these are square roots, you are looking for the largest factor of the radicand that is a perfect square.

For example, with $\sqrt{32ab^2}$ you are looking for the largest factor of $32ab^2$ that is a perfect square. To answer this question, you must first ask yourself, "What are the factors of $32$, $a$, and $b^2$?" Well, the factors of $32$ are $1, 2, 4, 8, 16,$ and $32$. The factors of $a$ are $1$ and $a$. The factors of $b^2$ are $1, b,$ and $b^2$.

To be even more precise, let's do an example. Say we wanted to simply $\sqrt{300ab^4}$. We would want to find the factors of $300$, $a$, and $b^4$ that are perfect squares. The largest factor of $300$ that is a perfect square is $100$. The largest factor of $a$ that is a perfect square is $1$. The largest factor of $b^4$ that is a perfect square is $b^4$ itself. Thus, the largest quantity we can take out of the radical is $100\cdot1\cdot b^4 = 100b^4$. The rest of the procedure goes like this:

$$ \sqrt{300ab^4} = \sqrt{100b^4 \cdot 3a} = \sqrt{100b^4}\sqrt{3a} = 10b^2\sqrt{3a}. $$

Try this process with the square roots you have in your post and see what happens.

As a side note, if you were instead trying to simplify the cube root of a quantity, you would look for the largest factor of the radicand that is a perfect cube, rather than a perfect square. With fourth roots, you'd look for perfect fourth powers, and so on.