Question

Find the constant term in (x/y -y/x)^12

Answer 1

Wolfram showed 924

Answer 2

The constant term would be when both (x/y) and (y/x) have the same power

Answer 3

do you know what power that would be?

Answer 4

the power would be 0

Answer 5

no

Answer 6

Each term is
\[ \binom{12}{r} \left ( \frac xy \right )^r \left ( - \frac yx \right )^{12-r}\]

Answer 7

wher r=0 to 12

Answer 8

:0 why?

Answer 9

Use the hint given by experimentX

Answer 10

yes, i agree with experimentX

Answer 11

so when will:

r = 12 - r

Answer 12

I must make use of the exponents?

Answer 13

when r =6?

Answer 14

r is the exponent of (x/y) and (12-r) is the exponent of (-y/x) --> when will these be equal?
yes 6 - that is correct

Answer 15

now use the formula given by experimentX to work out the answer

Answer 16

oooh...ok, the only thing i dont get is how do you know that r=12-r yields a constant term...

Answer 17

because:\[(\frac{x}{y})^6\times(\frac{-y}{x})^6\]will cancel all x and y variables

Answer 18

a bit more general form
\[ \binom{12}{r} (-1)^{12-r }\left ( \frac xy \right )^r \left (\frac yx \right )^{12-r} \]

Answer 19

wouldnt having any exponent that is the same for both terms cancel x and y too?

Answer 20

yes

Answer 21

and the only place in the expansion of that equation where both exponents are the same is at r=6

Answer 22

ooooooh, so would you generally use r= n-r; n is the exponent, to see this?

Answer 23

in cases like this yes.

Answer 24

only in cases with 2 different variables?

Answer 25

no - only in case where the product of the two variables cancels all "variable"s out.

Answer 26

so, its more of something that one must think about further?

Answer 27

(a-b)^n

if a*b results in a constant term then the above rule applies

Answer 28

what would be the other cases?

Answer 29

cases where a and b is not symmetric ..... though they might not always ask you to fing the coefficients where power is 0
they might ask you like
what is the coefficint of (x/y) .. etc etc

Answer 30

would \[x ^{3}((x ^{3}/y^2) + (3y/x^2))^9\] be a similar case?

Answer 31

I'm thinking not..? it doesnt look symmetrical...

Answer 32

we are looking for ?? with no x and y's??

Answer 33

yes, the constant term too

Answer 34

we are looking for 1/x^3

Answer 35

why is that? will this make the things cancel out too?

Answer 36

\[ \left ( \frac{x^3}{y^2} \right)^r \left ( \frac{3y}{x^2} \right)^{9-r} = \frac{1}{x^3}
\]

find the value of r

Answer 37

3r+9-r=0?

Answer 38

Oh sorry ... i meant 3

Answer 39

im not sure what to do now?

Answer 40

9-r-2r=0
3r - 2(9-r) = -3
solve for r

Answer 41

r=15?

Answer 42

it should be less than 9 and must be positive

Answer 43

oh wait, my bad, its 3

Answer 44

Umm ... i guess Ohh ...
if you cannot find +ve integer value less than .. the given power then you must conclude no solution exist.
I hope you can find it's coefficients then

Answer 45

i think i give up. Its unlikely that will come up in the test anyways. Thank you for your time and help:) much appreciated! :)

Answer 46

yw ... it's simple!!
r = 3
your answer is
\[ \binom{9}{3}(3)^{9-3}\]