If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer

$$a+b+c=4\,\Rightarrow\,(a+b+c)^2=16$$
$$(a^2+b^2+c^2)+2(ab+bc+ac)=16\Rightarrow2(ab+bc+ac)=9$$
Now:
$$(a+b+c)^3=(a^3+b^3+c^3)+3\left[(a+b+c)(ab+ac+bc)-abc\right]$$
$$64=28+3\left[4\times9-abc\right]\therefore\,abc=24$$
Now you need to try and simplify:
$${256=(a^4+b^4+c^4)+6(a^2b^2+a^2c^2+b^2c^2)+4(ab^3+ac^3+ba^3+bc^3+ca^3+cb^3)+12(abc^2+acb^2+bca^2)}$$
By using other methods like the fact that:
$$(a^2+b^2+c^2)^2=(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+a^2c^2)$$
