solve b-c=1; 2c3a=14; a-2b-4
Answers
Answer:
9
Step-by-step explanation:
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