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Question Number 152663 by mr W last updated on 31/Aug/21

$$\mathrm{If}{\mathrm{x}}^3-\mathrm{x}+3=0\mathrm{has}\mathrm{the}\mathrm{roots}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}.$$$$\mathrm{determine}\mathrm{the}\mathrm{monic}\mathrm{polynomial}\mathrm{with}$$$$\mathrm{the}\mathrm{roots}{\mathrm{a}}^5,{\mathrm{b}}^5\mathrm{and}{\mathrm{c}}^5.$$$$\left[Q152396\right]$$

Answered by mr W last updated on 30/Aug/21

$$a+b+c=0$$$$ab+bc+ca=-1$$$$abc=-3$$$$$$$$let{p}_k=a^k+b^k+c^k$$$$p_1=e_1=0$$$$p_2=e_1{p}_1-2e_2=-2\times (-1)=2$$$$p_3=e_1{p}_2-e_2{p}_1+3e_3=3(-3)=-9$$$$p_4=e_1{p}_3-e_2{p}_2+e_3{p}_1=-(-1)2=2$$$$p_5=e_1{p}_4-e_2{p}_3+e_3{p}_2=-(-1)(-9)+(-3)2=-15$$$$i.e.a^5+b^5+c^5=-15$$$$$$$$let{p}_k={\left(ab\right)}^k+{\left(bc\right)}^k+{\left(ca\right)}^k$$$$p_1=e_1=ab+bc+ca=-1$$$$e_2={ab}^{2}c+{bc}^{2}a+{ca}^{2}b=abc(a+b+c)=0$$$$e_3=ab\times bc\times ca={\left(abc\right)}^2=9$$$$p_2=e_1{p}_1-2e_2=(-1)(-1)-2\times 0=1$$$$p_3=e_1{p}_2-e_2{p}_1+3e_3=(-1)1+3\times 9=26$$$$p_4=e_1{p}_3-e_2{p}_2+e_3{p}_1=(-1)26+9(-1)=-35$$$$p_5=e_1{p}_4-e_2{p}_3+e_3{p}_2=(-1)(-35)+9\times 1=44$$$$i.e.a^5{b}^5+b^5{c}^5+c^5{a}^5=44$$$$$$$$a^5{b}^5{c}^5={\left(abc\right)}^5={(-3)}^5=-243$$$$$$$$theequationwithroots{a}^5,b^5,c^{5}is$$$$x^3+15x^2+44x+243=0$$

Commented by Tawa11 last updated on 30/Aug/21

$$\mathrm{Great}\mathrm{sir}$$

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