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The Particular and the General Case


Date: 11/06/2001 at 23:31:32
From: Phil Chui
Subject: Advanced Algebra

If a+b+c=3 and
   a(^2)+b(^2)+c(^2)=5 and
   a(^3)+b(^3)+c(^3)=7

what is a(^4)+b(^4)+c(^4) ?  And how is it done?


Date: 11/08/2001 at 17:12:29
From: Doctor Greenie
Subject: Re: Advanced Algebra

Hello, Phil -

I have seen problems like this before, but I have never taken the time 
to try to finish working one of them out to the end. Having had some 
free time since I first saw this question of yours, I have solved this 
problem - not just your specific case, but the general case. Thanks 
for sending the question; it gave me a lot of good (and enjoyable) 
mental exercise.

I'll present my solution roughly as I was able to figure it out. The 
actual path I took to the solution was more convoluted than the 
presentation below; I have cleaned things up a bit.

We start with

(1) a+b+c = 3
(2) a^2+b^2+c^2 = 5
(3) a^3+b^3+c^3 = 7

And we want to find the numerical value of

    a^4+b^4+c^4 = ?

I first noticed that I could get an expression including the required 
terms a^4, b^4, and c^4 by multiplying together either equations (1) 
and (3) above, or by multiplying equation (2) above by itself. I 
actually started down both paths more or less in parallel and chose 
the latter path when it appeared to hold more promise than the former.

So we have

   (a^2+b^2+c^2)^2 = (a^4+b^4+c^4)+2(a^2b^2+a^2c^2+b^2c^2)

and so

   (a^4+b^4+c^4) = (a^2+b^2+c^2)^2 - 2(a^2b^2+a^2c^2+b^2c^2)

Then, substituting from equation (2), we have

(4) (a^4+b^4+c^4) = 25 - 2(a^2b^2+a^2c^2+b^2c^2)

Now, to get a numerical value for (a^4+b^4+c^4), we need to evaluate 
the expression

   (a^2b^2+a^2c^2+b^2c^2)

After some pondering, I determined that I could obtain an expression 
including these terms by squaring the expression

   (ab+ac+bc)

and that, in turn, I could obtain an expression including these terms 
by squaring the given equation (1).

Note that I had no idea at this point whether this approach would lead 
to expressions that I could evaluate using equations (1), (2), and (3) 
- but, as you will see, it works out very nicely.

   (a+b+c)^2 = (a^2+b^2+c^2)+2(ab+ac+bc)

and so

   (ab+ac+bc) = [(a+b+c)^2 - (a^2+b^2+c^2)]/2

Then, substituting from equations (1) and (2), we have

(5)  (ab+ac+bc) = (9-5)/2 = 2

Next

   (ab+ac+bc)^2 = (a^2b^2+a^2c^2+b^2c^2)+2(a^2bc+ab^2c+abc^2)

and so

   (a^2b^2+a^2c^2+b^2c^2) = (ab+ac+bc)^2 - 2(a^2bc+ab^2c+abc^2)
                          = (ab+ac+bc)^2 - 2abc(a+b+c)

Then, substituting from equations (1) and (5), we have

(6)  (a^2b^2+a^2c^2+b^2c^2) = 2^2 - 2abc(3) = 4 - 6abc

And substituting (6) in (4), we now have

    (a^4+b^4+c^4) = 25 - 2(4 - 6abc)

or

(7) (a^4+b^4+c^4) = 17 + 12abc

So now we can evaluate the desired expression (a^4+b^4+c^4) if we can 
evaluate the expression abc.

When I got to this point, I realized I could get an expression 
involving the term abc by multiplying equation (1) by itself three 
times....

   (a+b+c)^3 = (a+b+c)(a+b+c)^2
             = (a+b+c)(a^2+b^2+c^2+2ab+2ac+2bc)
             = a^3+ ab^2+ ac^2+2a^2b+2a^2c+2abc
                              + a^2b      +2abc+b^3+ bc^2+2b^2c
                  +2ab^2+2ac^2      + a^2c+2abc    +2bc^2+ b^2c+c^3
              -----------------------------------------------------
             = a^3+3ab^2+3ac^2+3a^2b+3a^2c+6abc+b^3+3bc^2+3b^2c+c^3

             = (a^3+b^3+c^3)+3(ab^2+ac^2+a^2b+bc^2+a^2c+b^2c)+6abc

and so

   6abc = (a+b+c)^3 - (a^3+b^3+c^3) - 3(ab^2+ac^2+a^2b+bc^2+a^2c+b^2c)

Looking at this, I first tried grouping some terms...

6abc = (a+b+c)^3 - (a^3+b^3+c^3) - 3[a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)]

and then, after some examination of this expression, I saw that I 
could get clever by adding and subtracting 3(a^3+b^3+c^3) to the 
expression on the right:

   6abc = (a+b+c)^3 - (a^3+b^3+c^3) + 3(a^3+b^3+c^3)
          - 3[a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)] - 3(a^3+b^3+c^3)
   6abc = (a+b+c)^3 + 2(a^3+b^3+c^3)
          - 3[a(a^2+b^2+c^2)+b(a^2+b^2+c^2)+c(a^2+b^2+c^2)]

(8) 6abc = (a+b+c)^3 + 2(a^3+b^3+c^3)
          - 3(a+b+c)(a^2+b^2+c^2)

Substituting from equations (1), (2), and (3), we have

   6abc = 3^3 + 2(7) - 3(3)(5) = 27 + 14 - 45 = -4

and so

(9)  abc = -4/6 = -2/3

Then, finally, substituting this in equation (7), we have

   (a^4+b^4+c^4) = 17 + 12abc = 17 + 12(-2/3) = 17-8

and we finally have our result:

    a^4+b^4+c^4 = 9

********************************************

After going through the algebra for your particular case, I went back 
and worked out the general case:

(1)       a+b+c = x
(2) a^2+b^2+c^2 = y
(3) a^3+b^3+c^3 = z

I will spare you the details of the algebra for this general case (if 
you really love algebra, you might want to try to work it through for 
yourself). I came up with the following expression for a^4+b^4+c^4:

  a^4+b^4+c^4 = y^2 - 2[((x^2-y)/2)^2 - (x^4-3x^2y+2xz)/3]

I checked this result using the values from your problem. With x=3, 
y=5, and z=7, we get

  a^4+b^4+c^4 = 25 - 2[((9-5)/2)^2 - (3^4-3(3^2)(5)+2(3)(7))/3]
              = 25 - 2[4 - (81-135+42)/3]
              = 25 - 2[4 - (-12/3)]
              = 25 - 2(4+4)
              = 25 - 2(8)
              = 25 - 16
              = 9

You can also check the general result by choosing numbers for 
a, b, and c. For example, if a = 1, b = 2, c = 3, then x = a+b+c = 6, 
y = a^2+b^2+c^2 = 14, and z = a^3+b^3+c^3 = 36; the formula should 
give us a^4+b^4+c^4 = 81+16+1 = 98.

  a^4+b^4+c^4 = y^2 - 2[((x^2-y)/2)^2 - (x^4-3x^2y+2xz)/3]
              = 196 - 2[((36-14)/2)^2 - (1296-3(36)(14)+2(6)(36))/3]
              = 196 - 2[121 - (1296-1512+432)/3]
              = 196 - 2[121 - 216/3)
              = 196 - 2(121-72)
              = 196 - 2(49)
              = 196 - 98
              = 98

Thanks again for the nice problem.

Write back if you have any questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra

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