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### Topic: fulll hard question  (Read 19172 times)

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#### cuongt

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##### fulll hard question
« on: August 09, 2006, 07:34:44 AM »
the sum of three numbers is 4, the sum of their squares is 10 and the sum of their cubes is 22. What is the sum of their fourth powers ?

how the hek can u figure it out?

#### wereworm73

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##### Re: fulll hard question
« Reply #1 on: August 09, 2006, 04:30:18 PM »

#### lemonoman

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##### Re: fulll hard question
« Reply #2 on: August 09, 2006, 06:04:22 PM »
I had trouble following the instructions on wereworm's link...maybe my mind is just frozen.

But take a look

a + b + c = 4
a2 + b2 + c2 = 10
a3 + b3 + c3 = 21

3 equations, 3 unknowns.  Should be able to solve for a, b, and c and find the sum of the fourth powers directly.

#### Borek

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##### Re: fulll hard question
« Reply #3 on: August 09, 2006, 06:26:46 PM »
3 equations, 3 unknowns.  Should be able to solve for a, b, and c and find the sum of the fourth powers directly.

In general you are right about 3/3 - but two of the equations are not linear, so solving the set is tricky. And you are not guaranteed to find a real (ie no imaginary part) solution.
« Last Edit: August 09, 2006, 06:45:18 PM by Borek »
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#### xiankai

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##### Re: fulll hard question
« Reply #4 on: August 09, 2006, 07:06:04 PM »
thanks to the hint i managed to get an answer of 50. (not sure if its right, but i double checked my working already)

i was sort of stuck at the hard part of finding P  , until the hint came along

from this part onwards it should be easy
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#### lemonoman

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##### question
« Reply #5 on: August 09, 2006, 07:07:33 PM »
In general you are right about 3/3 - but two of the equations are not linear, so solving the set is tricky. And you are not guaranteed to find a real (ie no imaginary part) solution.

Oh yeah, well we'll just see!

a + b + c = 4
a2 + b2 + c2 = 10
a3 + b3 + c3 = 21

So c3 = 21 - a3 - b3
so c = (21 - a3 - b3)^(1/3)

So a2 + b2 + (21 - a3 - b3)^(2/3) = 10
and a + b + (21 - a3 - b3)^(1/3) = 4
this means (4 - a - b)^2 = 10 - a2 - b2
so 16 - 4a - 4b - 4a + a2 + ab - 4b + ab + b2 = 10 - a2 - b2
6 - 8a - 8b + 2a2 + 2b2 + 2ab = 0
(a2 - 4a) + (b2 - 4b) + ab = -3

Bah, you're right.

#### Borek

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##### Re: question
« Reply #6 on: August 10, 2006, 03:25:58 AM »
Bah, you're right.

You started wrong way - you should start with a = 4 - b - c, instert is into quadratic and cubic equation, then solve quadratic for b (or c) - and insert results (two of them) into cubic... This way it is doable, although you end with the cubic equation for c (or b).

I have never learnt any method of solving cubic equations, even if Junior showed me one last year and I have at least one book with other method on the shelf near my desk. But it is out of reach, as it is on the wrong end of the shelf
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#### FeLiXe

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##### Re: fulll hard question
« Reply #7 on: August 10, 2006, 07:21:13 AM »
you don't solve cubic equations with a formula

but you can always do it numerically = trying what works
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#### Borek

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##### Re: fulll hard question
« Reply #8 on: August 10, 2006, 08:31:57 AM »
you don't solve cubic equations with a formula

but you can always do it numerically = trying what works

You are not exactly right.

http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method
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#### sdekivit

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##### Re: fulll hard question
« Reply #9 on: August 10, 2006, 09:09:16 AM »
you don't solve cubic equations with a formula

but you can always do it numerically = trying what works

Cardano  but this is for equations like ax^3 + bx^2 + cx + d

#### Will

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##### Re: fulll hard question
« Reply #10 on: August 10, 2006, 09:16:29 AM »
It took me all morning it but it was worth it!
a = 2       _                      _
b = 1 + -/2   &   c = 1 - -/2

so xiankai is right that the sum of their fourth powers is 50.

#### Borek

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##### Re: fulll hard question
« Reply #11 on: August 10, 2006, 09:47:33 AM »
Cardano  but this is for equations like ax^3 + bx^2 + cx + d

Once you will go through the route described earlier (a = 4 - b - c inserted into quadratic, quadratic solved for - say - b - and this b inserted into cubic) you end with cubic equation in one variable (c in this case). That's where Cardano method can be used.
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#### sdekivit

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##### Re: fulll hard question
« Reply #12 on: August 10, 2006, 02:52:41 PM »
Cardano  but this is for equations like ax^3 + bx^2 + cx + d

Once you will go through the route described earlier (a = 4 - b - c inserted into quadratic, quadratic solved for - say - b - and this b inserted into cubic) you end with cubic equation in one variable (c in this case). That's where Cardano method can be used.

ok i haven't done that so i'll believe you on that

#### pantone159

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##### Re: fulll hard question
« Reply #13 on: August 10, 2006, 06:31:17 PM »
I got 50 as well, though I wasn't able to figure out a,b,c independently, other than they weren't all integers.

I did come up with the relations,
3 = ab + ac + bc
-2 = abc

Mostly my technique was to do things like
16 = (a+b+c)(a+b+c), expand the right side, and look for terms that matched the other two equations (e.g. aa+bb+cc=10.)
Enough grinding around with this and I got an answer.

#### xiankai

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##### Re: fulll hard question
« Reply #14 on: August 11, 2006, 05:56:32 AM »
Mostly my technique was to do things like
16 = (a+b+c)(a+b+c), expand the right side, and look for terms that matched the other two equations (e.g. aa+bb+cc=10.)
Enough grinding around with this and I got an answer.

exactly what i done too

i'll try that cardano method later, it looks promising.
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