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Algebra

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23 Feb 2008 09:57:23 IST

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491

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If a, b and c are the roots of the equation

x³-mx+19 = 0, find a³+b³+c^3.

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Comments (8)

Avi .....

Blazing goIITian

Joined: 19 Feb 2008

Posts: 318

23 Feb 2008 10:02:33 IST

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is it -57??

Hari Shankar

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Joined: 28 Feb 2007

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23 Feb 2008 10:03:29 IST

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method please!

Avi .....

Blazing goIITian

Joined: 19 Feb 2008

Posts: 318

23 Feb 2008 10:18:48 IST

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from thoery of roots we get
a+b+c=0
ab+bc+ac=-m
abc=-19

now

a³+ b³ + c³ - 3abc = (a+b+c)(a² +b² + c² -ab -bc -ac)

since a+b+c=0

a³ + b³ + c³= 3abc = 3*-19= -57

Avi .....

Blazing goIITian

Joined: 19 Feb 2008

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23 Feb 2008 10:19:12 IST

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rate if satisfied.....

cheeeerzz

narayana

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Joined: 13 Dec 2006

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23 Feb 2008 10:20:17 IST

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hey man it is -57

Hari Shankar

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Joined: 28 Feb 2007

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23 Feb 2008 10:45:46 IST

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Ok now I change the equation to

x³-5x²+8x-2 = 0

Now find a³+b³+c³

Avi .....

Blazing goIITian

Joined: 19 Feb 2008

Posts: 318

23 Feb 2008 10:49:53 IST

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now it comes as 11

Avi .....

Blazing goIITian

Joined: 19 Feb 2008

Posts: 318

23 Feb 2008 10:52:06 IST

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now
a+b+c = 5
ab+bc+ac = 8
abc = 2

same as above

a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c² -ab-bc-ac)

a³+b³+c³ = 5(25-3*8)+6 = 11

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