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kreshnik
 Post subject: I need help badly !
PostPosted: Sun, 25 Dec 2011 12:21:11 UTC 
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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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Denis
 Post subject: Re: Linear Equations
PostPosted: Sun, 25 Dec 2011 12:29:32 UTC 
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Please start your own thread...see "newtopic" button?

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outermeasure
 Post subject: Re: I need help badly !
PostPosted: Sun, 25 Dec 2011 12:54:42 UTC 
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Topic split.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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outermeasure
 Post subject: Re: I need help badly !
PostPosted: 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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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Denis
 Post subject: Re: I need help badly !
PostPosted: Sun, 25 Dec 2011 18:24:27 UTC 
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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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Denis
 Post subject: Re: I need help badly !
PostPosted: 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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kreshnik
 Post subject: Re: I need help badly !
PostPosted: Sun, 25 Dec 2011 18:55:45 UTC 
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Thanks a lot my friend...I appreciate that.
Thanks again :)
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Denis
 Post subject: Re: I need help badly !
PostPosted: Sun, 25 Dec 2011 19:03:08 UTC 
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kreshnik wrote:
Thanks a lot my friend...

THANK YOU: I had no friends before :oops:

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mathematic
 Post subject: Re: I need help badly !
PostPosted: Sun, 25 Dec 2011 23:05:42 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..

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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Shadow
 Post subject: Re: I need help badly !
PostPosted: Mon, 26 Dec 2011 00:38:16 UTC 
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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 x^2+y^2 from xy and (x+y)^2, then you can easily do the second one using x^3+y^3=(x+y)(x^2-xy+y^2).

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