The Particular and the General CaseDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/