Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$ How can I solve this question with out find a,b,c,d
$$a+b+c+d=2$$
$$a^2+b^2+c^2+d^2=30$$
$$a^3+b^3+c^3+d^3=44$$
$$a^4+b^4+c^4+d^4=354$$
so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$ 
If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d
Is there any help?

Here's an approach which involves AM-GM inequality(To bound the numbers).
I have considered the positive values of $(a,b,c,d)$. I can consider $a^2,b^2,c^2$ and $d^2$, since they are all strictly positive.
$\dfrac{a^2+b^2+c^2+d^2}{4} \ge \sqrt{abcd}$
$\dfrac{30}{4} \ge \sqrt{abcd} \implies 56.25 \ge abcd$
Since $a^2+b^2+c^2+d^2=30$ and $56 >|abcd|$, One of the $|a|,|b|,|c|,|d|$ is at most $5$.
Considering $a^4+b^4+c^4+d^4$, one of the $|a|,|b|,|c|,|d|$ is atmost $4$, since $5^4=625$ .
Now we have $a^2+b^2+c^2+d^2=30$, the value of $(|a|,|b|,|c|,|d|)$ is from set  $(1,2,3,4)$.
And also:
Considering $4th$ degree polynomial such that $a,b,c,d$ are the roots of the equation.
$x^4+px^3+qx^2+rx+s=0$
$a+b+c+d=2=-p$
$(a+b+c+d)^2-a^2+b^2+c^2+d^2=2(ab+bc+cd+ad+ac+bd)=q=-13$
Finding $\sum abc$ and $abcd$ with the help of other inequalities and finding the roots of equation is another way to go. (Not quite sure whether constructing polynomial is a great way to go. )

Edit: I have considered the absolute values of $(a,b,c,d)$ . I didn't assume them to be $+ve$. 
