imgur: the simple image sharer

I am having a tough time simplifying radicals after I multiply... Any help would be great, I already know the answer; I need to understand though, my textbook is not clear enough for me.

In the picture, you will see I wrote the answer down, but I couldn't finish my work because I did not know how.

it is c) e) and f)

It is very difficult to read what you have written. Is what looks like $2^4\sqrt{48}$ supposed to be $2\sqrt[4]{48}$?
If so then the problem is to write $2\sqrt[4]{48}\times \sqrt[4]{5}$ in the simplest form. The easiest way to do this, especially if you have "a tough time simplifying radicals after I multiply" is to simplify the radicals before you multiply. Here, recognise that 48= 2(24)= 2(2)(12)= 2(2)(2)(6)= 2(2)(2)(2)(2)(3) which is $2^4(3)$ so that $\sqrt[4]{48}= 2\sqrt[4]{3}$. So $2\sqrt{48}\times \sqrt[4]{5}= 2(2\sqrt[4]{3})(\sqrt[4]{5})= 4\sqrt{15}$

In (e), is that $\sqrt[3]{54 y^7}\times \sqrt[3]{6y^4}$? 3 divides into 7 twice with remainder 1 so $y^7= (y^3)(y^3)y$ so $\sqrt[3]{y^7}= y^2\sqrt[3]{y}$. 54= (27)(2) and $27= 3^3$ so $\sqrt[3]{54}= 3\sqrt[3]{2}$ and $\sqrt[3]{54y^7}= 3y^2\sqrt{2y}$. Of course, $y^4= y^3(y)$ while 6= 2(3) does not have any "cubes" so $\sqrt[3]{6y^4}= y\sqrt[3]{6y}$. So $\sqrt[3]{54y^7}\times\sqrt[3]{6y^4}= \left(3y^2\sqrt[3]{2y}\right)\left(y\sqrt{6y}\rignt)= 2y^3\sqrt[3]{12y^2}$.

f?

I'll post it here, though it also applies to #7. As HallsofIvy pointed out there is a problem with your notation. For example in f), what is under the radical in the first term. It looks like
$\sqrt{6}t \left ( 3t^2 \sqrt{\frac{t}{4}} \right )$

or
$\sqrt{6t} \left ( 3t^2 \frac{\sqrt{t}}{4} \right )$

or something else? Could you please insert some parenthesis in there?

-Dan

It is the first one; how do you guys type like that?