BRAIN BUSTER maths Q....?

Question

a+b+c=4
a^2+b^2+c^2=10
a^3+b^3+c^3=22
find the values of a, b ,and c

+say how you worked it out....
thanks !!!!

Answer 1

This doesn't answer your question, but I'm guessing that is very hard to do.
There is a very famous problem which goes something like this:
a+b+c = 4
a^2 + b^2 + c^2 = 10
a^3 + b^3 + c^3 = 22
Find a^4 + b^4 + c^4.

Now that is a much much easier problem. If you want a solution to that, let me know, and I'll do it for you. I have a good feeling thats what you wanted.. if not, I apologise.

Answer 2

Sorry Scott you dont have to be that rude! I did this work and you can see that the way you did it was different from my way! If you get the 10 points is fine I mean, you were the first one... but you have no right to insulte my work and call it plagium because math problems have many way to be solved and I'm sure you know that!!!

Lets name each equation as 1, 2 and 3

Now I am going to work with eq 3

a^3+b^3+c^3=22

a^3 + b^3 = 22 - c^3

Factorizing...

(a+b) (a^2 -ab +b^2) = 22 - c^3

But in eq 1 we now that a+b= 4-c and
also in eq 1
(a+b+c)^2 = 16
a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 16
10 + 2ab + 2ac + 2bc = 16 (subs eq 2 here)
ab + ac + bc = 3
ab = 3 -c (a+b)
ab = 3 -c (4-c)
ab= 3 -4c - c^2

Now we back to (a+b) (a^2 -ab +b^2) = 22 - c^3
and we sustitute all the work we did

(4-c) (10 - c^2 - c^2 +4c -3) = 22 - c^3

(c-4) (2c^2 -4c -7) = (22-c^3)

c^3 -4c^2 +3c + 2= 0

OK I CONTINUE!!!

factorizing that equation we have

(c-2) (c^2 -2c -1) = 0

and we have c = 2, c= 1 + sqroot(2) and c= 1 - sqroot (2)

And if we sustitute c = 2 in the first eq we have a+b=2
and in 2nd eq a^2 + b^2 = 6
a=2-b
a^2 = b^2 -4b +4

sustiute that in eq resulting of the 2nd eq

b^2 -4b +4 + b^2 = 6

2b^2 - 4b - 2 = 0

b^2 -2b - 1 = 0

we have b = 1 + sqroot (2) and b = 1 - sqroot (2)

we get the same results for a... so we find this combinantions

( 1 + sqroot (2), 1 - sqroot (2), 2)
(1 - sqroot (2), 1 + sqroot (2), 2)
(2, 1 + sqroot (2), 1 - sqroot (2))
(2, 1 - sqroot (2), 1 + sqroot (2))
(1 + sqroot (2), 2, 1 - sqroot (2))
(1 - sqroot (2), 2, 1 + sqroot (2))

DONE !!!!

Answer 3

6 solutions for {a,b,c}:

{ 2, 1-√2, 1+√2 }
{ 2, 1+√2, 1-√2 }
{ 1+√2, 2, 1-√2 }
{ 1+√2, 1-√2, 2 }
{ 1-√2, 1+√2, 2 }
{ 1-√2, 2, 1+√2 }
****************************** ******************************
PROOF:
4 = a+b+c is given.

16 = (a+b+c)^2 = [a^2+b^2+c^2] + 2(ab+ac+bc) = 10 + 2(ab+ac+bc)
so 2(ab + ac + bc) = 6
or ab+ac+bc = 3

64 = (a+b+c)^3 = [a^3+b^3+c^3] + 3(a+b+c)(ab+ac+bc) - 3abc
or
64 = 22 + 3*4*3 - 3abc
64 = 22 + 36 - 3abc
6 = -3abc
or abc = -2

Now we have:

a+b+c = 4 or b+c = 4-a
abc = -2 or bc = -2/a
ab+ac+bc = 3 or a(b+c) + bc = 3

Substituting the first two into the third:

a(4-a) -2/a = 3
or
a^3 - 4a^2 + 3a + 2 = 0

leading to three values for a:
a = 2
a = 1-√2
a = 1+√2

By symmetry when a takes one of these, b takes one of the other two and c takes the last, leading to the 6 permutations given above.

Q.E.D.

Good problem.
Oh, and a^4 + b^4 + c^4 = 50 too.

Answer 4

I doubt that anyone will be able to give you a good answer for this one. The methods for solving cubic polynomials are difficult to use, tend to give you answers that involve very large formulae and it is often difficult to confirm that these answers are correct (if one makes a mistake applying the formulae, and it is easy to do so, it would be easy to miss that mistake). You'll be able to find an approximate solution on something like Maple or Matlab. If someone can see a solution, I'm happily proved wrong.