Question

2 cosx = x + (1/x) then value of x^6 - (1/x^6) is?

Answer 1

find the value of
x^3 + 1/x^3 and x^3 - 1/x^3

Answer 2

\[z+\frac{1}{z}=2\cos x\]\[z^6-\frac{1}{z^6}=?\]

Answer 3

square it ... try some manipulation ...
( x - (1/x))^2 +2 = ?
a^3 + b^3 = ?
a^3 - b^3 = ?
make use of these basic simple formulas.

Answer 4

or one can write\[z=e^{ix}\]and find the value of\[z^6-z^{-6}\]

Answer 5

nice observation ... didn't think of that.

Answer 6

@mukushla i cant see ur...Witings....

Answer 7

reload the page

Answer 8

lol....i mean....Font is Small

Answer 9

lol\[\large z=e^{ix}\]find the value of\[\large z^6-z^{-6}\]

Answer 10

e^ix - 1/e^ix

Answer 11

\[\large z^6-z^{-6}=e^{6xi}-e^{-6xi}=\cos 6x+i\sin 6x-(\cos 6x-i\sin 6x)\]is this right?
i have some doubts on it

Answer 12

Yes...u r correct

Answer 13

so it gives answer as \[2i \sin 6x\]