+8262 
What do you call an insect that plays drums?
how do you solve them? Please help. Just solve the first one step by step. Thank you.
+564 Okay solved another one just now after class. Some of these are kind of tricky so I'll solve some of them when I'm a little more awake xD
$$\frac{a^{3}-49a}{a^{3}+7a^{2}}$$
$$=\frac{(a-7)(a^{2}+7a)}{a^{3}+7a^{2}}$$
$$=\frac{(a-7)(a^{2}+7a)}{a(a^{2}+7a)}$$ cross out the $$(a^{2}+7a)$$ from the numerator and denominator...
$$=\frac{a-7}{a}$$ and here is your final answer!
------
Now time to do my Political Science homeowork xD
+11912 i remember this!my drama teacher told me ,................its a rhythma tick , the answer!
+118687 $$\dfrac{6a^5b^4}{9a^3b^7}\\\\
=\dfrac{2a^5b^4}{3a^3b^7}\qquad $ I have cancelled top and bottom by 3$\\\\
=\dfrac{2aaaaabbbb}{3aaabbbbbbb}\qquad $ I have expanded, there are invisable multiply signs between the letters$\\\\
=\dfrac{2aabbbb}{3bbbbbbb} \qquad $ I have cancelled out aaa top and bottom $\\\\
=\dfrac{2aa}{3bbb} \qquad $ I have cancelled out bbbb top and bottom $\\\\
=\dfrac{2a^2}{3b^3}$$
+8262 Ok. Thank you. Now that I have an idea, do I solve the second one step by step?
+118687 Yes.
You may be able to do more than one cancellation at the same time but it might be better to do it slowly like I have until you get the hang of it. 
+8262 Ok here it goes!
$$\frac{12a^5b^3(3-b)}{4a^4b(b^2+b-12)}\\\\=\frac{3a^5b^3(3-b)}{1a^4b(b^2b-12)}\mbox{I\;have\;canceled\;top\;and\;bottom\;by\;4.}\\\\=\frac{3aaaaabbb(3-b)}{4aaaab(bb+b-12)}\mbox{Now\;that\;I\;have\;expanded,\;there\;are\;no\;invisible\;signs\;between\;the\;letters.}\\\\=\frac{3abbb(3-b)}{4b(bb+b-12)}\mbox{I\;have\;cancelled\;out\;}aaaa\;\mbox{top\;and\;bottom.}\\\\=\frac{3abb(3-b)}{4(bb+b-12)}\mbox{I\;have\;cancelled\;}b\;\mbox{top\;and\;bottom.}\\\\=\frac{3abb(3-b)}{4(bb+b-12)}$$
+118687 Is this really your homework Dragon.
Most of what you have dones really good but you made a mistake with the numbers 4divided by 4=1
Sp there is no 4 on the bottom.
After this it gets far to difficult for you!!!
$$\\\frac{3abb(3-b)}{1(bb+b-12)}\\\\
=\frac{3ab^2(3-b)}{b^2+b-12}\\\\
$ and that is as far as your teacher could expect you to get - the rest is FAR too difficult. I'll finish if for you just this time$\\\\
=\frac{-3ab^2(b-3)}{b^2+b-12}\\\\
=\frac{-3ab^2(b-3)}{(b+4)(b-3)}\\\\
=\frac{-3ab^2}{b+4}\\\\
=-\frac{3ab^2}{b+4}\\\\$$
+8262 So do I do the top one only? Yes. The teacher have us this as homework to PREPARE US FOR 10TH GRADE.
+564 I'll try and help you with a few. This thing looks like a lot of work lol .-.
+8262 Oh! Thank god you came! Thank you very much! I just needed someone to help. Thank you!
+8262 I need help with the ones of adding, subtracting, or known as the long ones.
I appreciate your help.
+564 Okay, I solved one this morning before I went to school:
$$\frac{(b-5)^{3}}{15+7b-2b^{2}}$$
$$\frac{(b-5)(b-5)(b-5)}{(-2b-3)(b-5)}$$
Cancel like terms (the (b-5) at the top and bottom)
$$-\frac{(b-5)^{2}}{(2b-3)}$$ This is your final answer :D
+564 Okay solved another one just now after class. Some of these are kind of tricky so I'll solve some of them when I'm a little more awake xD
$$\frac{a^{3}-49a}{a^{3}+7a^{2}}$$
$$=\frac{(a-7)(a^{2}+7a)}{a^{3}+7a^{2}}$$
$$=\frac{(a-7)(a^{2}+7a)}{a(a^{2}+7a)}$$ cross out the $$(a^{2}+7a)$$ from the numerator and denominator...
$$=\frac{a-7}{a}$$ and here is your final answer!
------
Now time to do my Political Science homeowork xD
+8262 Today. I will ask the teacher how to do this, simulating that I don't know.
