5. Add a b (b + c) +bc(6-) and 2b (5-0) - be (a + b).
Answers
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.