## GoldRush18 4 years ago 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

1. GoldRush18

this is a Hard one for me

2. experimentX

??? $x^4 - \frac{y^4}{x-y}$

3. GoldRush18

let me try to write that one better hold on

4. experimentX

$\frac{x^4 - y^4}{x-y}$

5. GoldRush18

$x^4-y^4\div x-y$ Simplify

6. experimentX

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

7. GoldRush18

well this question has 3 parts to it

8. anonymous

you got this?

9. GoldRush18

nope i dont fully understand it

10. 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

11. anonymous

now replace $$x$$ by $$y+1$$

12. experimentX

LOL ... i forgot the question!!

13. anonymous

denominator is $$y+1-y=1$$ numerator is $(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)$

14. GoldRush18

oh ok

15. 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?

16. 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

17. GoldRush18

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

18. anonymous

yes

19. GoldRush18

thats what i see on the paper and i double checked

20. GoldRush18

either way i think its correct nonetheless

21. anonymous

$(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)$ $=((y+1)+y)((y+1)^2+y^2)$ $=(y+1)^3+y(y+1)^2+y^2(y+1)+y^2$

22. GoldRush18

is this for the third part

23. anonymous

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

24. GoldRush18

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

25. 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

26. GoldRush18

deduce that $(y+1)^4-y^4<4(y+1)^3$

27. experimentX

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

28. anonymous

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

29. GoldRush18

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

30. 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

31. experimentX

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

32. experimentX

y^2(y+1) < (y+1)^3 so, $(y+1)^3+y(y+1)^2+y^2(y+1)+y^2 < (y+1)^3 + (y+1)^3 + (y+1)^3 + (y+1)^3 = 4(y+1)^3$

33. experimentX

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

34. experimentX

so, $(y+1)^4 - y^4 < 4(y+1)^3$

35. GoldRush18

OK i get it now

36. GoldRush18

thanks @satellite73 and @experimentX :)

37. experimentX

yw!!

38. 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

39. 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

40. 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

41. experimentX

you mean equation 2 ??

42. 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

43. 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!!

44. 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

45. GoldRush18

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

46. anonymous

nvm

47. 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!!

48. anonymous

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