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

- GoldRush18

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- GoldRush18

this is a Hard one for me

- experimentX

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

- GoldRush18

let me try to write that one better hold on

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## More answers

- experimentX

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

- GoldRush18

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

- experimentX

(x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2)
I think you can do the rest!!!

- GoldRush18

well this question has 3 parts to it

- anonymous

you got this?

- GoldRush18

nope i dont fully understand it

- anonymous

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

- anonymous

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

- experimentX

LOL ... i forgot the question!!

- anonymous

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

- GoldRush18

oh ok

- anonymous

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?

- anonymous

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

- GoldRush18

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

- anonymous

yes

- GoldRush18

thats what i see on the paper and i double checked

- GoldRush18

either way i think its correct nonetheless

- anonymous

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

- GoldRush18

is this for the third part

- anonymous

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

- GoldRush18

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

- anonymous

@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

- GoldRush18

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

- experimentX

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

- anonymous

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

- GoldRush18

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

- anonymous

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

- experimentX

y^3<(y+1)^3
y(y+1)^2<(y+1)^3
so on ..

- experimentX

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

- experimentX

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

- experimentX

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

- GoldRush18

OK i get it now

- GoldRush18

thanks @satellite73 and @experimentX :)

- experimentX

yw!!

- anonymous

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

- anonymous

@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

- anonymous

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

- experimentX

you mean equation 2 ??

- anonymous

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

- experimentX

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!!

- anonymous

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

- GoldRush18

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

- anonymous

nvm

- experimentX

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

- anonymous

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

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