Thread: Any kind hearted soul, radical expressions help!!!

Please help me with these two questions. I have a major headache from doing it!

Hint- . Also, .

(57) looks fairly straight-forward. Have you tried multiplying? Show us your effort and share your thoughts, and we can better help you.

The second problem gives you a hint. What did you get?

$(x^2 + \sqrt{2}xy + y^2) * (x^2 - \sqrt{2}xy + y^2) = what?$

By analogy then $x^8 + y^8 = what?$

Thanks for the reply! As you can see in my image, for question 53, i cannot find any perfect squares in the square root. I do not know how to advance from there on.

As for question 57, how do I go on from there? I do see some similarities with X^4 and Y^4 in my half-completed answer. However, how do I remove the 2x^2y^2-2xy that u see in my answer ?

$\sqrt{x^{2n}} * \sqrt{x^3y^{3n}} * \sqrt{y^{(n+1)}} = \sqrt{x^{2n} * x^3 * y^{3n} * y^{(n + 1)}}= \sqrt{x^{(2n + 3)} * y^{(4n + 1)}}.$ So far so good.

But $2n + 3 = (2n + 2) + 1 = 2(n + 1) + 1.$ So

$\sqrt{x^{(2n + 3)} * y^{4n + 1}} = \sqrt{x^{\{2(n + 1) + 1\}} * y^{(4n+1)}} = \sqrt{x^{\{2(n + 1)\}} * x^1 * y^{4n} * y^1} = x^{(n + 1)} * y^{2n} * \sqrt{xy}.$

As for the second problem, you did the algebra wrong

$(x^2 + \sqrt{2}xy + y^2)(x^2 - \sqrt{2}xy + y^2) = x^4 -\sqrt{2}x^3y + x^2y^2 + \sqrt{2}x^3y - 2x^2y^2 + \sqrt{2}xy^3 + x^2y^2 -\sqrt{2}xy^3 + y^4 =$

$x^4 -\sqrt{2}x^3y + \sqrt{2}x^3y + x^2y^2 + x^2y^2 - 2x^2y^2 + \sqrt{2}xy^3 -\sqrt{2}xy^3 + y^4 = x^4 + y^4.$

Now think about $x^8 + y^8.$

hey JeffM. Isn't Squareroot2xy multiplied by squareroot2xy = 2xy? as you remove the squares with the square root.

Originally Posted by xwy

hey JeffM. Isn't Squareroot2xy multiplied by squareroot2xy = 2xy? as you remove the squares with the square root.

You are correct that: $\sqrt{2xy} * \sqrt{2xy} = \sqrt{2xy * 2xy} = \sqrt{2^2x^2y^2} = 2xy,\ assuming\ xy \ge 0.$

But if I understood your problem, that is NOT what you are dealing with.

$\sqrt{2}xy * \sqrt{2}xy = \sqrt{2} * xy * \sqrt{2} * xy = \sqrt{2} * \sqrt{2} * x^2y^2 = \sqrt{2 * 2} * x^2y^2 = \sqrt{4} * x^2y^2 = 2x^2y^2.$

If I have led you astray by not understanding (or by making ANOTHER idiotic mistake), I do apologize.

Hey no worries. I really appreciate your help!

This post should be deleted as soon as the preceding post is, but why anyone would pay money for academic help to someone who cannot bother to proof his text for gross errors in English grammar and spelling is beyond me. I guess it is true that a fool and his money are soon parted.

Originally Posted by JeffM

This post should be deleted as soon as the preceding post is, but why anyone would pay money for academic help to someone who cannot bother to proof his text for gross errors in English grammar and spelling is beyond me. I guess it is true that a fool and his money are soon parted.

I agree with you entirely. However responding to such simply gives the spammer a larger chance of someone seeing his/her post, which is the whole idea. Please report the post and either mash or I will get to it as soon as possible.

-Dan