Question

let x and y be positive real numbers such that x does not equal to y.
i. simplify x^4-y^4/ x-y
ii. hence, or otherwise, show that
(y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3.
iii. Deduce that (y+1)^4-y^4 < 4(y+1)^3

Answer 1

this is a Hard one for me

Answer 2

??? \[ x^4 - \frac{y^4}{x-y}\]

Answer 3

let me try to write that one better hold on

Answer 4

\[ \frac{x^4 - y^4}{x-y}\]

Answer 5

\[x^4-y^4\div x-y\] Simplify

Answer 6

(x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2)

I think you can do the rest!!!

Answer 7

well this question has 3 parts to it

Answer 8

you got this?

Answer 9

nope i dont fully understand it

Answer 10

\[\frac{x^4 - y^4}{x-y}=\frac{(x+y)(x-y)(x^2+y^2)}{x-y}=(x+y)(x^2+y^2)\] is the first part

Answer 11

now replace \(x\) by \(y+1\)

Answer 12

LOL ... i forgot the question!!

Answer 13

denominator is \(y+1-y=1\) numerator is
\[(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)\]

Answer 14

oh ok

Answer 15

it is not clear to me what you wrote for the second part, but now algebra should give us what want. what was the second part supposed to be?

Answer 16

you wrote
\[(y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3.\] but the parentheses are messed up

Answer 17

so replace x=y+1 into the first question?

Answer 18

yes

Answer 19

thats what i see on the paper and i double checked

Answer 20

either way i think its correct nonetheless

Answer 21

\[(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)\]
\[=((y+1)+y)((y+1)^2+y^2)\]
\[=(y+1)^3+y(y+1)^2+y^2(y+1)+y^2\]

Answer 22

is this for the third part

Answer 23

i do not get what you want for the second part. there is a hanging plus sign and an extraneous )

Answer 24

i got the second part alright its the third part i don't get let me rewrite it here

Answer 25

@experimentX do you know what the second part is supposed to be? it makes no sense to me. there is a typo of some sort

Answer 26

deduce that \[(y+1)^4-y^4<4(y+1)^3\]

Answer 27

sorry .. i was busy somewhere!!
i'll see

Answer 28

so far i have this
\[(y+1)^4-y^4=(y+1)^3+y(y+1)^2+y^2(y+1)+y^3\]

Answer 29

the second one to me is just simply saying the answer u got in first one just replace x=y+1 thats all

Answer 30

yes, but it has a specific form it wants you to put it in, and i don't know what that form is supposed to be

Answer 31

y^3<(y+1)^3
y(y+1)^2<(y+1)^3

so on ..

Answer 32

y^2(y+1) < (y+1)^3
so,
\[ (y+1)^3+y(y+1)^2+y^2(y+1)+y^2 < (y+1)^3 + (y+1)^3 + (y+1)^3 + (y+1)^3 = 4(y+1)^3\]

Answer 33

looks like equation went off screen
\[(y+1)^3+y(y+1)^2+y^2(y+1)+y^2 < 4(y+1)^3 \]

Answer 34

so,
\[ (y+1)^4 - y^4 < 4(y+1)^3\]

Answer 35

OK i get it now

Answer 36

thanks @satellite73 and @experimentX :)

Answer 37

yw!!

Answer 38

maybe it is supposed to be this:
\[(y+1)^4-y^4=(y+1)^3+y(y+1)^2+y^2(y+1)+y^3\]
well i don't

Answer 39

@experimentX do you have any idea what the right side is supposed to be in equation 2?
i mean i get the inequality, just not what the expression is supposed to be

Answer 40

this line \[(y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3\] has no meaning to me

Answer 41

you mean equation 2 ??

Answer 42

i got this
\[(y+1)^4-y^4=(y+1)^3+y(y+1)^2+y^2(y+1)+y^3\]
, but it is not clear to me what we are supposed to turn it in to

Answer 43

i think there a typho there!!
a^n - b^n = (a - n) (a^(n-1) + a^(n-2)b + a^(n-3)b^2 + ... +b^(n-1))
this is the pattern!!

Answer 44

ok i got that. i won't worry about it
i know there is a typo but i am not sure what it was supposed to be instead

Answer 45

maybe there was a typo because in maths alot of errors are made

Answer 46

nvm

Answer 47

\( (y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3. \)
^

\( (y+1)^4-y^4=(y+1)^3+(y+1)^2+(y+1)y^2+y^3. \)
^

also,
\( (y+1)^4-y^4=(y+1)^3+(y+1)^2+(y+1)y^2+y^3. \)
^ <---- y missing here .... i think ti should be what you got!!

Answer 48

got it. backwards parentheses and missing y confused the heck out of me