Evaluating the value of the roots of the expression












0












$begingroup$


I'm attempting this question below, I understand the method when its squared but not cubed. I evaluated $a^2 + b^2 + c^2$ as $(a + b + c)^2 -2(ab + ac + bc)$. I tried to obtain the answer by guess and check, changing a few of the powers around to try and work backwards but to no success. The answer for the question below is 23, and based upon the responses provided $a^4 + b^4 + c^4$ should be a breeze also? Thanks



Q: $x^3 - 2x^2 - 3x + 1 = 0$



$a^3 + b^3 +c^3 =?$






polynomials roots







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    a,b,c are the roots, right?
    $endgroup$
    – Hello_World
    Jan 6 at 0:12










  • $begingroup$
    Yes, apologies, I added a y instead of a c above.
    $endgroup$
    – Zach
    Jan 6 at 0:13
















0












$begingroup$


I'm attempting this question below, I understand the method when its squared but not cubed. I evaluated $a^2 + b^2 + c^2$ as $(a + b + c)^2 -2(ab + ac + bc)$. I tried to obtain the answer by guess and check, changing a few of the powers around to try and work backwards but to no success. The answer for the question below is 23, and based upon the responses provided $a^4 + b^4 + c^4$ should be a breeze also? Thanks



Q: $x^3 - 2x^2 - 3x + 1 = 0$



$a^3 + b^3 +c^3 =?$






polynomials roots







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    a,b,c are the roots, right?
    $endgroup$
    – Hello_World
    Jan 6 at 0:12










  • $begingroup$
    Yes, apologies, I added a y instead of a c above.
    $endgroup$
    – Zach
    Jan 6 at 0:13














0












0








0


1



$begingroup$


I'm attempting this question below, I understand the method when its squared but not cubed. I evaluated $a^2 + b^2 + c^2$ as $(a + b + c)^2 -2(ab + ac + bc)$. I tried to obtain the answer by guess and check, changing a few of the powers around to try and work backwards but to no success. The answer for the question below is 23, and based upon the responses provided $a^4 + b^4 + c^4$ should be a breeze also? Thanks



Q: $x^3 - 2x^2 - 3x + 1 = 0$



$a^3 + b^3 +c^3 =?$






polynomials roots







share|cite|improve this question











$endgroup$




I'm attempting this question below, I understand the method when its squared but not cubed. I evaluated $a^2 + b^2 + c^2$ as $(a + b + c)^2 -2(ab + ac + bc)$. I tried to obtain the answer by guess and check, changing a few of the powers around to try and work backwards but to no success. The answer for the question below is 23, and based upon the responses provided $a^4 + b^4 + c^4$ should be a breeze also? Thanks



Q: $x^3 - 2x^2 - 3x + 1 = 0$



$a^3 + b^3 +c^3 =?$






polynomials roots




polynomials roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 0:13







Zach

















asked Jan 6 at 0:11









ZachZach

365




365








  • 1




    $begingroup$
    a,b,c are the roots, right?
    $endgroup$
    – Hello_World
    Jan 6 at 0:12










  • $begingroup$
    Yes, apologies, I added a y instead of a c above.
    $endgroup$
    – Zach
    Jan 6 at 0:13














  • 1




    $begingroup$
    a,b,c are the roots, right?
    $endgroup$
    – Hello_World
    Jan 6 at 0:12










  • $begingroup$
    Yes, apologies, I added a y instead of a c above.
    $endgroup$
    – Zach
    Jan 6 at 0:13








1




1




$begingroup$
a,b,c are the roots, right?
$endgroup$
– Hello_World
Jan 6 at 0:12




$begingroup$
a,b,c are the roots, right?
$endgroup$
– Hello_World
Jan 6 at 0:12












$begingroup$
Yes, apologies, I added a y instead of a c above.
$endgroup$
– Zach
Jan 6 at 0:13




$begingroup$
Yes, apologies, I added a y instead of a c above.
$endgroup$
– Zach
Jan 6 at 0:13










3 Answers
3






active

oldest

votes


















1












$begingroup$

I suggest that you check this out: https://en.m.wikipedia.org/wiki/Newton%27s_identities



Now, the idea, as you guess, is to write $a^3+b^3+c^3$ as a function of the symmetric elementary polynomials.



Now, $a^3=2a^2+3a-1$, same for $b,c$.



Thus $a^3+b^3+c^3=2(a^2+b^2+c^2)+3(a+b+c)-3$, hence
$a^3+b^3+c^3=2(a+b+c)^2+3(a+b+c)-4(ab+bc+ca)-3$.



Finally, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $$x^3+1=2x^2+3x$$



    Cubing both sides $$(x^3+1)^3=(2x^2+3x)^3=(2x^2)^3+(3x)^3+3(2x^2)(3x)(2x^2+3x)$$



    Setting $x^3=y$ $$(y+1)^3=8y^2+27y+18y(y+1)$$



    $$iff y^3+y^2(3-8-18)+y(cdots)+cdots=0$$ whose roots are $a^3,b^3,c^3$



    $$a^3+b^3+c^3=dfrac{23}1$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      By Vieta's formulas:
      $$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
      Then you can use the formulas:
      $$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
      &=2^2-2(-3)=\
      &=10.end{align}$$
      $$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
      &=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
      &=2(2^2-3(-3))+3(-1)=\
      &=23.end{align}$$
      $$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
      &=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
      &=10^2-2[(-3)^2-2(-1)2]=\
      &=126.end{align}$$
      $$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
      & 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
      &=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
      &=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
      &=2^5-frac{5[-15]}{3}(7)=\
      &=207.end{align}$$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
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        3 Answers
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        active

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        1












        $begingroup$

        I suggest that you check this out: https://en.m.wikipedia.org/wiki/Newton%27s_identities



        Now, the idea, as you guess, is to write $a^3+b^3+c^3$ as a function of the symmetric elementary polynomials.



        Now, $a^3=2a^2+3a-1$, same for $b,c$.



        Thus $a^3+b^3+c^3=2(a^2+b^2+c^2)+3(a+b+c)-3$, hence
        $a^3+b^3+c^3=2(a+b+c)^2+3(a+b+c)-4(ab+bc+ca)-3$.



        Finally, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          I suggest that you check this out: https://en.m.wikipedia.org/wiki/Newton%27s_identities



          Now, the idea, as you guess, is to write $a^3+b^3+c^3$ as a function of the symmetric elementary polynomials.



          Now, $a^3=2a^2+3a-1$, same for $b,c$.



          Thus $a^3+b^3+c^3=2(a^2+b^2+c^2)+3(a+b+c)-3$, hence
          $a^3+b^3+c^3=2(a+b+c)^2+3(a+b+c)-4(ab+bc+ca)-3$.



          Finally, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            I suggest that you check this out: https://en.m.wikipedia.org/wiki/Newton%27s_identities



            Now, the idea, as you guess, is to write $a^3+b^3+c^3$ as a function of the symmetric elementary polynomials.



            Now, $a^3=2a^2+3a-1$, same for $b,c$.



            Thus $a^3+b^3+c^3=2(a^2+b^2+c^2)+3(a+b+c)-3$, hence
            $a^3+b^3+c^3=2(a+b+c)^2+3(a+b+c)-4(ab+bc+ca)-3$.



            Finally, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.






            share|cite|improve this answer









            $endgroup$



            I suggest that you check this out: https://en.m.wikipedia.org/wiki/Newton%27s_identities



            Now, the idea, as you guess, is to write $a^3+b^3+c^3$ as a function of the symmetric elementary polynomials.



            Now, $a^3=2a^2+3a-1$, same for $b,c$.



            Thus $a^3+b^3+c^3=2(a^2+b^2+c^2)+3(a+b+c)-3$, hence
            $a^3+b^3+c^3=2(a+b+c)^2+3(a+b+c)-4(ab+bc+ca)-3$.



            Finally, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 0:20









            MindlackMindlack

            3,81518




            3,81518























                1












                $begingroup$

                $$x^3+1=2x^2+3x$$



                Cubing both sides $$(x^3+1)^3=(2x^2+3x)^3=(2x^2)^3+(3x)^3+3(2x^2)(3x)(2x^2+3x)$$



                Setting $x^3=y$ $$(y+1)^3=8y^2+27y+18y(y+1)$$



                $$iff y^3+y^2(3-8-18)+y(cdots)+cdots=0$$ whose roots are $a^3,b^3,c^3$



                $$a^3+b^3+c^3=dfrac{23}1$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  $$x^3+1=2x^2+3x$$



                  Cubing both sides $$(x^3+1)^3=(2x^2+3x)^3=(2x^2)^3+(3x)^3+3(2x^2)(3x)(2x^2+3x)$$



                  Setting $x^3=y$ $$(y+1)^3=8y^2+27y+18y(y+1)$$



                  $$iff y^3+y^2(3-8-18)+y(cdots)+cdots=0$$ whose roots are $a^3,b^3,c^3$



                  $$a^3+b^3+c^3=dfrac{23}1$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    $$x^3+1=2x^2+3x$$



                    Cubing both sides $$(x^3+1)^3=(2x^2+3x)^3=(2x^2)^3+(3x)^3+3(2x^2)(3x)(2x^2+3x)$$



                    Setting $x^3=y$ $$(y+1)^3=8y^2+27y+18y(y+1)$$



                    $$iff y^3+y^2(3-8-18)+y(cdots)+cdots=0$$ whose roots are $a^3,b^3,c^3$



                    $$a^3+b^3+c^3=dfrac{23}1$$






                    share|cite|improve this answer









                    $endgroup$



                    $$x^3+1=2x^2+3x$$



                    Cubing both sides $$(x^3+1)^3=(2x^2+3x)^3=(2x^2)^3+(3x)^3+3(2x^2)(3x)(2x^2+3x)$$



                    Setting $x^3=y$ $$(y+1)^3=8y^2+27y+18y(y+1)$$



                    $$iff y^3+y^2(3-8-18)+y(cdots)+cdots=0$$ whose roots are $a^3,b^3,c^3$



                    $$a^3+b^3+c^3=dfrac{23}1$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 6 at 10:15









                    lab bhattacharjeelab bhattacharjee

                    225k15157275




                    225k15157275























                        0












                        $begingroup$

                        By Vieta's formulas:
                        $$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
                        Then you can use the formulas:
                        $$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
                        &=2^2-2(-3)=\
                        &=10.end{align}$$
                        $$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
                        &=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
                        &=2(2^2-3(-3))+3(-1)=\
                        &=23.end{align}$$
                        $$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
                        &=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
                        &=10^2-2[(-3)^2-2(-1)2]=\
                        &=126.end{align}$$
                        $$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
                        & 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
                        &=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
                        &=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
                        &=2^5-frac{5[-15]}{3}(7)=\
                        &=207.end{align}$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          By Vieta's formulas:
                          $$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
                          Then you can use the formulas:
                          $$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
                          &=2^2-2(-3)=\
                          &=10.end{align}$$
                          $$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
                          &=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
                          &=2(2^2-3(-3))+3(-1)=\
                          &=23.end{align}$$
                          $$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
                          &=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
                          &=10^2-2[(-3)^2-2(-1)2]=\
                          &=126.end{align}$$
                          $$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
                          & 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
                          &=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
                          &=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
                          &=2^5-frac{5[-15]}{3}(7)=\
                          &=207.end{align}$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            By Vieta's formulas:
                            $$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
                            Then you can use the formulas:
                            $$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
                            &=2^2-2(-3)=\
                            &=10.end{align}$$
                            $$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
                            &=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
                            &=2(2^2-3(-3))+3(-1)=\
                            &=23.end{align}$$
                            $$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
                            &=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
                            &=10^2-2[(-3)^2-2(-1)2]=\
                            &=126.end{align}$$
                            $$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
                            & 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
                            &=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
                            &=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
                            &=2^5-frac{5[-15]}{3}(7)=\
                            &=207.end{align}$$






                            share|cite|improve this answer









                            $endgroup$



                            By Vieta's formulas:
                            $$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
                            Then you can use the formulas:
                            $$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
                            &=2^2-2(-3)=\
                            &=10.end{align}$$
                            $$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
                            &=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
                            &=2(2^2-3(-3))+3(-1)=\
                            &=23.end{align}$$
                            $$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
                            &=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
                            &=10^2-2[(-3)^2-2(-1)2]=\
                            &=126.end{align}$$
                            $$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
                            & 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
                            &=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
                            &=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
                            &=2^5-frac{5[-15]}{3}(7)=\
                            &=207.end{align}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 6 at 11:56









                            farruhotafarruhota

                            20.2k2738




                            20.2k2738






























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                                0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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                                張江高科駅

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                                張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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