GoldRush18
  • GoldRush18
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
GoldRush18
  • GoldRush18
this is a Hard one for me
experimentX
  • experimentX
??? \[ x^4 - \frac{y^4}{x-y}\]
GoldRush18
  • GoldRush18
let me try to write that one better hold on

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

experimentX
  • experimentX
\[ \frac{x^4 - y^4}{x-y}\]
GoldRush18
  • GoldRush18
\[x^4-y^4\div x-y\] Simplify
experimentX
  • experimentX
(x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2) I think you can do the rest!!!
GoldRush18
  • GoldRush18
well this question has 3 parts to it
anonymous
  • anonymous
you got this?
GoldRush18
  • GoldRush18
nope i dont fully understand it
anonymous
  • 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
  • anonymous
now replace \(x\) by \(y+1\)
experimentX
  • experimentX
LOL ... i forgot the question!!
anonymous
  • anonymous
denominator is \(y+1-y=1\) numerator is \[(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)\]
GoldRush18
  • GoldRush18
oh ok
anonymous
  • 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
  • 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
  • GoldRush18
so replace x=y+1 into the first question?
anonymous
  • anonymous
yes
GoldRush18
  • GoldRush18
thats what i see on the paper and i double checked
GoldRush18
  • GoldRush18
either way i think its correct nonetheless
anonymous
  • 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
  • GoldRush18
is this for the third part
anonymous
  • anonymous
i do not get what you want for the second part. there is a hanging plus sign and an extraneous )
GoldRush18
  • GoldRush18
i got the second part alright its the third part i don't get let me rewrite it here
anonymous
  • 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
  • GoldRush18
deduce that \[(y+1)^4-y^4<4(y+1)^3\]
experimentX
  • experimentX
sorry .. i was busy somewhere!! i'll see
anonymous
  • 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
  • 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
  • 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
  • experimentX
y^3<(y+1)^3 y(y+1)^2<(y+1)^3 so on ..
experimentX
  • 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
  • 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
  • experimentX
so, \[ (y+1)^4 - y^4 < 4(y+1)^3\]
GoldRush18
  • GoldRush18
OK i get it now
GoldRush18
  • GoldRush18
thanks @satellite73 and @experimentX :)
experimentX
  • experimentX
yw!!
anonymous
  • 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
  • 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
  • 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
  • experimentX
you mean equation 2 ??
anonymous
  • 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
  • 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
  • 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
  • GoldRush18
maybe there was a typo because in maths alot of errors are made
anonymous
  • anonymous
nvm
experimentX
  • 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
  • anonymous
got it. backwards parentheses and missing y confused the heck out of me

Looking for something else?

Not the answer you are looking for? Search for more explanations.