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 Post subject: I need help badly !Posted: Sun, 25 Dec 2011 12:21:11 UTC
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Joined: Sun, 25 Dec 2011 00:28:25 UTC
Posts: 95
what should I do with these??!

1. x^3-y^3=28 and x-y=4, then ... x^2+xy+y^2=?

2. x+y=9 , x*y=8 then ... x^3+y^3=?

sorry for interrupting..

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 Post subject: Re: Linear EquationsPosted: Sun, 25 Dec 2011 12:29:32 UTC
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 Post subject: Re: I need help badly !Posted: Sun, 25 Dec 2011 12:54:42 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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Topic split.

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 Post subject: Re: I need help badly !Posted: Sun, 25 Dec 2011 12:55:57 UTC
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kreshnik wrote:
what should I do with these??!

1. x^3-y^3=28 and x-y=4, then ... x^2+xy+y^2=?

2. x+y=9 , x*y=8 then ... x^3+y^3=?

sorry for interrupting..

What have you tried?

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 Post subject: Re: I need help badly !Posted: Sun, 25 Dec 2011 18:24:27 UTC
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Joined: Sun, 24 Jul 2005 20:12:39 UTC
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Location: Ottawa Ontario
kreshnik wrote:
2. x+y=9 , x*y=8 then ... x^3+y^3=?

Your English is quite good; better than some around here!!
You got that one right. Good work.

May have been easier this way:
x + y = 9 ; y = 9 - x [1]
xy = 8 [2]
Substitute [1] in [2]:
x(9 - x) = 8
9x - x^2 = 8
x^2 - 9x + 8 = 0 ; factor:
(x - 8)(x - 1) = 0
x = 8 or x = 1
Substitute in [1]:
y = 9-8 = 1 or y = 9-1 = 8

x^3 + y^3 = 8^3 + 1^3 = 513
or
x^3 + y^3 = 1^3 + 8^3 = 513

That's just another way of doing it...

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 Post subject: Re: I need help badly !Posted: Sun, 25 Dec 2011 18:35:05 UTC
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kreshnik wrote:
1. x^3-y^3=28 and x-y=4, then ... x^2+xy+y^2=?

You were ok up to this line:
-3yxx + 3yyx = 64 - 28 ; then you somehow got your signs wrong; should be:

3xy^2 - 3yx^2 = 36 ; divide by 3:
xy^2 - yx^2 = 12

Substitute x = y + 4 (from the given x - y = 4):
y^2(y + 4) - y(y + 4)^2 = 12

I'll let you finish it...

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 Post subject: Re: I need help badly !Posted: Sun, 25 Dec 2011 18:55:45 UTC
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Joined: Sun, 25 Dec 2011 00:28:25 UTC
Posts: 95
Thanks a lot my friend...I appreciate that.
Thanks again

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 Post subject: Re: I need help badly !Posted: Sun, 25 Dec 2011 19:03:08 UTC
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Location: Ottawa Ontario
kreshnik wrote:
Thanks a lot my friend...

THANK YOU: I had no friends before

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 Post subject: Re: I need help badly !Posted: Sun, 25 Dec 2011 23:05:42 UTC
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Joined: Fri, 1 Jul 2011 01:17:26 UTC
Posts: 321
kreshnik wrote:
what should I do with these??!

1. x^3-y^3=28 and x-y=4, then ... x^2+xy+y^2=?

2. x+y=9 , x*y=8 then ... x^3+y^3=?

sorry for interrupting..

1. is very easy when you recognize x^3 - y^3 = (x - y)(x^2 + xy +y^2)

2. is slightly harder: (x + y)^3 = x^3 + 3yx^2 + 3xy^2 + y^3 = x^3 + y^3 + 3(x + y)xy.

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 Post subject: Re: I need help badly !Posted: Mon, 26 Dec 2011 00:38:16 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12101
Location: Austin, TX
mathematic wrote:
kreshnik wrote:
what should I do with these??!

1. x^3-y^3=28 and x-y=4, then ... x^2+xy+y^2=?

2. x+y=9 , x*y=8 then ... x^3+y^3=?

sorry for interrupting..

1. is very easy when you recognize x^3 - y^3 = (x - y)(x^2 + xy +y^2)

2. is slightly harder: (x + y)^3 = x^3 + 3yx^2 + 3xy^2 + y^3 = x^3 + y^3 + 3(x + y)xy.

They have the first one first for a reason, you realize how to get from and , then you can easily do the second one using .

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