Hint- $\displaystyle \sqrt{a} = a^{\frac{1}{2}}$. Also, $\displaystyle \sqrt{a} \sqrt{b} = \sqrt{ab}$.
(57) looks fairly straight-forward. Have you tried multiplying? Show us your effort and share your thoughts, and we can better help you.
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 $x^8 + y^8.$
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.
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.