Here is one problem.
If
$\displaystyle a+b+c=0$
and
$\displaystyle a^2+b^2+c^2=1$
calculate $\displaystyle a^4+b^4+c^4$
Help anyone?
Perhaps you may do this,Originally Posted by DenMac21
$\displaystyle a^2+b^2+c^2=1$
Non of the $\displaystyle a,b,c$ could be more than 1, thus, $\displaystyle |a,b,c|<1$
Do this in terms of a function of one, $\displaystyle a=x$ where $\displaystyle x$ is a changing variable with $\displaystyle |x|<1$. Thus,
$\displaystyle x+b+c=0$
$\displaystyle x^2+b^2+c^2=1$
Now solve these two for $\displaystyle b,c$ it does not matter because we can consider them to be the same here. Thus, by substitution,
$\displaystyle x^2+(-x-c)^2+c^2=1$ thus,
$\displaystyle x^2+x^2+2cx+c^2+c^2=1$ thus,
$\displaystyle x^2+cx+c^2=1/2$ thus,
$\displaystyle c^2+xc+(x^2-1/2)=0$ thus,
$\displaystyle c=\frac{-x\pm\sqrt{2-3x^2}}{2}$ by Quadradic Formula, thus,
$\displaystyle b=\frac{-x\mp\sqrt{2-3x^2}}{2}$
Thus, as a function of $\displaystyle x$ for $\displaystyle |x|<1$
$\displaystyle (a,b,c)=(x,\frac{-x\pm\sqrt{2-3x^2}}{2},\frac{-x\mp\sqrt{2-3x^2}}{2})$
You can check that,
$\displaystyle a+b+c=0$
$\displaystyle a^2+b^2+c^2=1$.
Know all is left is to find what,
$\displaystyle a^4+b^4+c^4$ as see whether it is invaraint of $\displaystyle x$.
Here it gets messy, notice that
$\displaystyle (n+m)^4+(n-m)^4=2n^4+12n^2m^2+m^4$
This is confirmed by simple expansion.
Thus,
$\displaystyle a^4+b^4+c^4$ becomes,
$\displaystyle x^4+\left(\frac{-x+\sqrt{2-3x^2}}{2}\right)^4+\left(\frac{-x-\sqrt{2-3x^2}}{2}\right)^4$
Now this is a variation of the identity before show, with
$\displaystyle n=-x,m=\frac{-x+\sqrt{2-3x^2}}{2}$
Thus,
$\displaystyle a^4+b^4+c^4=\frac{1}{2}$
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).