## GoldRush18 Group Title 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 2 years ago 2 years ago

1. GoldRush18 Group Title

this is a Hard one for me

2. experimentX Group Title

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

3. GoldRush18 Group Title

let me try to write that one better hold on

4. experimentX Group Title

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

5. GoldRush18 Group Title

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

6. experimentX Group Title

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

7. GoldRush18 Group Title

well this question has 3 parts to it

8. satellite73 Group Title

you got this?

9. GoldRush18 Group Title

nope i dont fully understand it

10. satellite73 Group Title

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

11. satellite73 Group Title

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

12. experimentX Group Title

LOL ... i forgot the question!!

13. satellite73 Group Title

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

14. GoldRush18 Group Title

oh ok

15. satellite73 Group Title

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. satellite73 Group Title

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 Group Title

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

18. satellite73 Group Title

yes

19. GoldRush18 Group Title

thats what i see on the paper and i double checked

20. GoldRush18 Group Title

either way i think its correct nonetheless

21. satellite73 Group Title

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

22. GoldRush18 Group Title

is this for the third part

23. satellite73 Group Title

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

24. GoldRush18 Group Title

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

25. satellite73 Group Title

@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 Group Title

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

27. experimentX Group Title

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

28. satellite73 Group Title

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 Group Title

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

30. satellite73 Group Title

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 Group Title

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

32. experimentX Group Title

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 Group Title

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 Group Title

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

35. GoldRush18 Group Title

OK i get it now

36. GoldRush18 Group Title

thanks @satellite73 and @experimentX :)

37. experimentX Group Title

yw!!

38. satellite73 Group Title

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. satellite73 Group Title

@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. satellite73 Group Title

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 Group Title

you mean equation 2 ??

42. satellite73 Group Title

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 Group Title

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. satellite73 Group Title

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 Group Title

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

46. satellite73 Group Title

nvm

47. experimentX Group Title

$$(y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3.$$ ^ $$(y+1)^4-y^4=(y+1)^3+(y+1)^2+(y+1)y^2+y^3.$$ ^ 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. satellite73 Group Title

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