Here is one problem.
If
and
calculate
Help anyone?
Perhaps you may do this,Originally Posted by DenMac21
Non of the could be more than 1, thus,
Do this in terms of a function of one, where is a changing variable with . Thus,
Now solve these two for it does not matter because we can consider them to be the same here. Thus, by substitution,
thus,
thus,
thus,
thus,
by Quadradic Formula, thus,
Thus, as a function of for
You can check that,
.
Know all is left is to find what,
as see whether it is invaraint of .
Here it gets messy, notice that
This is confirmed by simple expansion.
Thus,
becomes,
Now this is a variation of the identity before show, with
Thus,
Here is one way.Originally Posted by DenMac21
a +b +c = 0 ------------(1)
a^2 +b^2 +c^2 = 1 -----------(2)
a^2 +b^2 +c^2 = 1 ------------------------------------(2)
Square both sides,
(a^4 +b^4 +c^4) +2[(ab)^2 +(ac)^2 +(bc)^2] = 1 -------(2a)
So,
a^4 +b^4 +c^4 = 1 -2[(ab)^2 +(ac)^2 +(bc)^2] ----------(2b)
a +b +c = 0 -----------------------------(1)
Square both sides,
(a^2 +b^2 +c^2) +2[ab +ac +bc] = 0 ------(1a)
Substitution, from (2),
1 +2[ab +ac +bc] = 0
2[ab +ac +bc] = -1
ab +ac +bc = -1/2
Square both sides,
[(ab)^2 +(ac)^2 +(bc)^2] +2[(a^2)bc +a(b^2)c +ab(c^2)] = 1/4 -----(3)
[(ab)^2 +(ac)^2 +(bc)^2] +2abc[a +b +c] = 1/4
Since [a+b+c] = 0, then,
[(ab)^2 +(ac)^2 +(bc)^2] = 1/4 -----------***
Substitute that into (2b),
a^4 +b^4 +c^4 = 1 -2[1/4]
Therefore,
a^4 +b^4 +c^4 = 1/2 -------------answer.
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I did the long multiplications on paper here. I did not show them here because, well, they are long.
You can do them on paper there to verify how I got the Eq.(2a), (1a), and (3).