 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Summing powers (Posted on 2004-07-01) I thought of three numbers.
Their sum is 6.
The sum of their squares is 8.
The sum of their cubes is 5.
What is the sum of their fourth powers?

 See The Solution Submitted by Federico Kereki Rating: 3.0000 (5 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Imagine | Comment 8 of 18 | An imaginary number is x + y*i  where i= sqrt(-1)
The real part is x, the imaginary part is y
That number squared is (x^2 - y^2) + 2*x*y*i
... and  cubed is (x^3 - 3*x*y^2) + (3*x^2*y  - y^3)*i

a+bi, c+di, e+fi     so there are 6 variables

And we have 3 sums:
6+0i, 8+0i, 5+0i

So we can make 6 equations with 6 unknowns:
1 Sum Realpart N = 6
2 Sum Imaginary N =0
3 Sum Realpart Nsquared = 8
4 Sum Imaginary Nsquared =0
5 Sum Realpart Ncubed =5
6 Sum Imaginary Ncubed =0

1: a+c+e=6
2: b+d+f=0
3: a^2 - b^2 +c^2 - d^2 +e^2 - f^2 =8
4: 2*a*b +2*c*d +2*e*f =0
5: a^3 - 3*a*b^2 + c^3 - 3*c*d^2 + e^3 - 3*e*f^2 =5
6: 3*a^2*b - b^3 + 3*c^2*d  - d^3 + 3*e^2*f  - f^3 =0

So this suggests to me that there is a complex number solution.  Anyone care to solve these 6 equations?

 Posted by Larry on 2004-07-05 02:21:13 Please log in:

 Search: Search body:
Forums (2)