Question

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

Answer 1

know the a^3- b^3 formula?

Answer 2

yep

Answer 3

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

Answer 4

difference^2 = sum^2 -4*product

Answer 5

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.

Answer 6

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

Answer 7

for the second one?

Answer 8

second question?
how to do?

Answer 9

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.

Answer 10

@satellite73

Answer 11

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

Answer 12

@nincompoop please help

Answer 13

@dan815 please help

Answer 14

@phi please help

Answer 15

@Loser66 @Hero please help

Answer 16

@mukushla please help

Answer 17

what do you think about the first question? @Hero

Answer 18

sry Michele what's the problem?

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

the question is the same

Answer 23

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

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

thanks! for your replies @dan815

Answer 28

yep sure thing :)

Answer 29

thank you so much :)