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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

Mathematics
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this is a Hard one for me
??? \[ x^4 - \frac{y^4}{x-y}\]
let me try to write that one better hold on

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Other answers:

\[ \frac{x^4 - y^4}{x-y}\]
\[x^4-y^4\div x-y\] Simplify
(x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2) I think you can do the rest!!!
well this question has 3 parts to it
you got this?
nope i dont fully understand it
\[\frac{x^4 - y^4}{x-y}=\frac{(x+y)(x-y)(x^2+y^2)}{x-y}=(x+y)(x^2+y^2)\] is the first part
now replace \(x\) by \(y+1\)
LOL ... i forgot the question!!
denominator is \(y+1-y=1\) numerator is \[(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)\]
oh ok
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?
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
so replace x=y+1 into the first question?
yes
thats what i see on the paper and i double checked
either way i think its correct nonetheless
\[(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)\] \[=((y+1)+y)((y+1)^2+y^2)\] \[=(y+1)^3+y(y+1)^2+y^2(y+1)+y^2\]
is this for the third part
i do not get what you want for the second part. there is a hanging plus sign and an extraneous )
i got the second part alright its the third part i don't get let me rewrite it here
@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
deduce that \[(y+1)^4-y^4<4(y+1)^3\]
sorry .. i was busy somewhere!! i'll see
so far i have this \[(y+1)^4-y^4=(y+1)^3+y(y+1)^2+y^2(y+1)+y^3\]
the second one to me is just simply saying the answer u got in first one just replace x=y+1 thats all
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
y^3<(y+1)^3 y(y+1)^2<(y+1)^3 so on ..
y^2(y+1) < (y+1)^3 so, \[ (y+1)^3+y(y+1)^2+y^2(y+1)+y^2 < (y+1)^3 + (y+1)^3 + (y+1)^3 + (y+1)^3 = 4(y+1)^3\]
looks like equation went off screen \[(y+1)^3+y(y+1)^2+y^2(y+1)+y^2 < 4(y+1)^3 \]
so, \[ (y+1)^4 - y^4 < 4(y+1)^3\]
OK i get it now
yw!!
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
@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
this line \[(y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3\] has no meaning to me
you mean equation 2 ??
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
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!!
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
maybe there was a typo because in maths alot of errors are made
nvm
\( (y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3. \) ^ \( (y+1)^4-y^4=(y+1)^3+(y+1)^2+(y+1)y^2+y^3. \) ^ 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!!
got it. backwards parentheses and missing y confused the heck out of me

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