If α ,β ,γ are three numbers s.t.:
$\ α^ \ $ + $\ β \ $ + $ γ \ $ = −2
$\ α^2 \ $ + $\ β^2 \ $ + $ γ^2 \ $ = 6
$\ α^3 \ $ + $\ β^3 \ $ + $ γ^3 \ $ = −5,
then $\ α^4 \ $ + $\ β^4 \ $ + $ γ^4 \ $ is equal to ??
I tried out substituting the values of each equation to one other ...but it became very complex .. I also remember some crammers rule for this ..using matrices?? Is that the way??
Let $A_{n}=a^n+b^n+c^n$. Then we have $$A_{n+3}=(a+b+c)A_{n+2}-(ab+bc+ac)A_{n+1}+abcA_{n}$$ and $$2(ab+bc+ac)=(a+b+c)^2-(a^2+b^2+c^2)=4-6=-2$$
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$ Then we can easily find $abc$.
HINT:
$$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$$
Now $$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)$$
$$2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)$$ we can find $ab+bc+ca$ from here
$$a^3+b^3+c^3-3abc$$
$$=(a+b)^3-3ab(a+b)+c^3-3abc$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2+c^2-3ab\}$$
$$= (a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$$ we can find $abc$ from here
