+546 Hello!
So when i posted these previously, they were marked wrong by my intructor...she wants them done a certain way and says that the way CPhill was doing them were "overthought" :( sorry CPhill, you have helped so much! I know you can do them the way she wants, so do you think you view the page showing how she does them?
Combine radicals. Simplify in to perfect cube. Take cubed root.
30) √8x^5*√3x
32) 3 cubedsqrt5y^3* 2 cubedsqrt50y^4
38) √48x^3/√3xy^2
40) cubedsqrt250x^7y^3/cubedsqrt2x^2y
+118608 okay Natazsaa, that is a good idea.
I strongly suspect that all our answers are the same, just presented differently so that they look different.
You post your teacher's answer and tell us which of our answers it compares to.
Then we can look and maybe explain the confusion to you. 
+118608 30) √8x^5*√3x
$$\sqrt8x^5\sqrt3x = \sqrt{24}x^6 \\\\$$
There is no cubed root in this question. 
+33616 I guess 32) might be:
$$3\sqrt[3]{5y^3}\times 2\sqrt[3]{50y^4}$$
Multiply the terms outside the surds together, and multiply the terms inside the surds together:
$$6\times \sqrt[3]{250y^7}$$
Now you want to express the terms inside the surd as something-cubed multiplied by whatever is left over and can't be expressed as a nice cube!
$$250=2\times 5^3\\
y^7=y\times (y^2)^3$$
So:
$$6\sqrt[3]{250y^7}=6\sqrt[3]{2\times 5^3\times y\times (y^2)^3}=6\times5\times y^2\times \sqrt[3]{2y}$$
or just:
$$30y^2\sqrt[3]{2y}$$
But you must use brackets more to ensure your meaning is clear and unambiguous Nataszaa!
+118608 40) cubedsqrt250x^7y^3/cubedsqrt2x^2y
$$\frac{\sqrt[3]{250x^7y^3}}{\sqrt[3]{2x^2y}}\\\\
=\sqrt[3]{\frac{250x^7y^3}{2x^2y}}\\\\
=\sqrt[3]{\frac{250x^{7-1}y^{3}}{2xy}}\\\\
=\sqrt[3]{\frac{2\times 125 x^{6}y^{3}}{2xy}}\\\\
=\sqrt[3]{\frac{(5x^2y)^3}{xy}}\\\\
=\frac{\sqrt[3]{(5x^2y)^3}}{\sqrt[3]{xy}}}\\\\
=\dfrac{5x^2y}{\sqrt[3]{xy}}}\\\\$$
I did some 'repairs' after Nataszaa comment. Thankyou for pointing that out Natasza, I don't like leaving errors laying about!
+546 Hey Meoldy!
Not sure if you read my entire post...but are you doing these the same way that CPhill was? I totally appreciate his work, and he always gets answers right...but for some reason these square roots radicals have been wrong. im not sure about your answers...i will know in an hour
+118608 Nataszaa, why don't you post one of your teacher's answers (with the question)
CPhill is a very good mathematician, if his answer is incorrect it is most likely because you did not put brackets in and he has answered a question that is different from the one you intended. (I may well have done the same thing)
Maybe we can teach you how to present your questions so that they will not be misinterpreted.
+118608 38) √48x^3/√3xy^2
I have answered the question you have asked but I may not have answered the question that you intended.
$$\frac{\sqrt{48}x^3}{\sqrt3y^2}\\\\
=\frac{\sqrt{48/3}x^3}{y^2}\\\\
=\frac{\sqrt{16}x^3}{y^2}\\\\
=\frac{4x^3}{y^2}\\\\$$
+546 Hey Melody
I know that CPhill is AMAZING and he always does these questions for me...so i myself, was taken aback when she said they were wrong. i will post how SHE does them for ya.
+128591 cubedsqrt250x^7y^3/cubedsqrt2x^2y =
3√(250*x^7*y^3)/ 3√(2*x^2*y) = 3√(125*x^5*y^2) = (5x)[3√(x^2*y^2)]
+118608 okay Natazsaa, that is a good idea.
I strongly suspect that all our answers are the same, just presented differently so that they look different.
You post your teacher's answer and tell us which of our answers it compares to.
Then we can look and maybe explain the confusion to you.
