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 $a^5+b^5+c^5$

Question

I have tried to solve it but cannot find any approach which would lead me to the answer

Answer 1

Hint

Use Newton’s identities and Vieta's formulas to find $a,b,c$ as roots of a degree 3 polynomial.

Answer 2

Use $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$

and $$a^3+b^3+c^3-3bc=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$$ to find $ab+bc+ca=u, abc=v$(say)

So, $a,b,c$ are the roots of $$t^3-4t^2+ut-v=0$$

Use Newton's Sums

Answer 3

Maple:

simplify(a^4+b^4+c^4, {a+b+c = 4, a^2+b^2+c^2 = 7, a^3+b^3+c^3 = 28}) = 209/2

simplify(a^5+b^5+c^5, {a+b+c = 4, a^2+b^2+c^2 = 7, a^3+b^3+c^3 = 28}) = 334

Answer 4

$$a+b+c=4\,\Rightarrow\,(a+b+c)^2=16$$ $$(a^2+b^2+c^2)+2(ab+bc+ac)=16\Rightarrow2(ab+bc+ac)=9$$ 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: $$(a^2+b^2+c^2)^2=(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+a^2c^2)$$