## rvc one year ago Please Help. I need help with the first two questions. And is the third answer's solution's step is correct?

1. rvc

2. hartnn

know the a^3- b^3 formula?

3. rvc

yep

4. hartnn

use the property that, in a quadratic equation ax^2+bx+c=0 sum of roots = -b/a product of roots = c/a find the difference of roots from the above two values

5. hartnn

difference^2 = sum^2 -4*product

6. anonymous

The third one looks correct. So the solutions would be the appropriate roots of unity. For the second one , write $$x_n$$ as $$e^{i\frac{\pi}{2^n}}$$ and proceed.

7. rvc

hmmm.. i gtg now too busy atm ill check these sums later. Thanks to all. If i require help i will surely tag u ppl. :) thanks once again

8. rvc

for the second one?

9. rvc

second question? how to do?

10. anonymous

Like I said, note that $$x_n = e^{i\frac{\pi}{2^{n}}}$$ Now multiply the $$x_n$$'s and you will notice an infinite sum in the exponent. Tell me when you get that far.

11. rvc

@satellite73

12. Michele_Laino

please, note that, in the first question, we have not a quadratic equation @hartnn

13. Michele_Laino

14. Michele_Laino

15. Michele_Laino

16. Michele_Laino

17. Michele_Laino

18. Michele_Laino

what do you think about the first question? @Hero

19. anonymous

sry Michele what's the problem?

20. Michele_Laino

if please can help us to solve the first question @mukushla

21. anonymous

@rvc please re-check the problem 1, it seems incorrect!

22. dan815

try just plugging in a and b and get expressions for a and b

23. rvc

the question is the same

24. dan815

$a^2-6\sqrt{a}+4=0\\ b^2-6\sqrt{b}+4=0$

25. dan815

not sure where to take it from here u can get some expressions for a^3-b^3

26. dan815

$a^3=(-4+6\sqrt{a})^{3/2}\\ b^3=(-4+6\sqrt{b})^{3/2}$ or $(\frac{(a^2+4)}{6})^6=a^3 \\ (\frac{(b^2+4)}{6})^6=b^3$

27. dan815

oh i checked ur solution on wolfram i think your question is wrong, because http://www.wolframalpha.com/input/?i=x^2-6*sqrt%28x%29%2B4%3D0 that equation as real roots

28. Michele_Laino

29. dan815

yep sure thing :)

30. rvc

thank you so much :)