$p^3+q^3+r^3$, where $p$,$q$, and $r$ are the roots of the cubic function $x^3+4x^2-4x+1$. Working included.











up vote
0
down vote

favorite












I am trying to utilise the expressions for Vieta's fomulae to solve expressions, just as an investigation. The question I gave myself is, if $p$, $q$, and $r$ are the roots of a cubic, what is $p^3+q^3+r^3$? The cubic is $x^3+4x^2-4x+1.$



I have done all the working out and come to the conclusion that the answer is $-115$. However, I am not 100% sure that my working out is correct, or if I used the formulae in the right way. If someone could check my working that I have attached as an image and point out any fundamental errors, that would be great. Thanks!



WORKING HERE






roots







share|cite|improve this question
























  • There's no image. But in any case, you should use MathJax rather than an image.
    – Robert Israel
    Nov 29 at 3:57










  • I have updated the post. Also, there are many lines of working and I'm new to Maths Stack Exchange so I feel that in this moment in time neat, written working is the most appropriate way to go :)
    – Chem_Student
    Nov 29 at 4:00










  • The correct answer is $-115$.
    – Robert Israel
    Nov 29 at 4:01










  • See math.stackexchange.com/questions/3015661/…
    – lab bhattacharjee
    Nov 29 at 4:03















up vote
0
down vote

favorite












I am trying to utilise the expressions for Vieta's fomulae to solve expressions, just as an investigation. The question I gave myself is, if $p$, $q$, and $r$ are the roots of a cubic, what is $p^3+q^3+r^3$? The cubic is $x^3+4x^2-4x+1.$



I have done all the working out and come to the conclusion that the answer is $-115$. However, I am not 100% sure that my working out is correct, or if I used the formulae in the right way. If someone could check my working that I have attached as an image and point out any fundamental errors, that would be great. Thanks!



WORKING HERE






roots







share|cite|improve this question
























  • There's no image. But in any case, you should use MathJax rather than an image.
    – Robert Israel
    Nov 29 at 3:57










  • I have updated the post. Also, there are many lines of working and I'm new to Maths Stack Exchange so I feel that in this moment in time neat, written working is the most appropriate way to go :)
    – Chem_Student
    Nov 29 at 4:00










  • The correct answer is $-115$.
    – Robert Israel
    Nov 29 at 4:01










  • See math.stackexchange.com/questions/3015661/…
    – lab bhattacharjee
    Nov 29 at 4:03













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am trying to utilise the expressions for Vieta's fomulae to solve expressions, just as an investigation. The question I gave myself is, if $p$, $q$, and $r$ are the roots of a cubic, what is $p^3+q^3+r^3$? The cubic is $x^3+4x^2-4x+1.$



I have done all the working out and come to the conclusion that the answer is $-115$. However, I am not 100% sure that my working out is correct, or if I used the formulae in the right way. If someone could check my working that I have attached as an image and point out any fundamental errors, that would be great. Thanks!



WORKING HERE






roots







share|cite|improve this question















I am trying to utilise the expressions for Vieta's fomulae to solve expressions, just as an investigation. The question I gave myself is, if $p$, $q$, and $r$ are the roots of a cubic, what is $p^3+q^3+r^3$? The cubic is $x^3+4x^2-4x+1.$



I have done all the working out and come to the conclusion that the answer is $-115$. However, I am not 100% sure that my working out is correct, or if I used the formulae in the right way. If someone could check my working that I have attached as an image and point out any fundamental errors, that would be great. Thanks!



WORKING HERE






roots




roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 3:58

























asked Nov 29 at 3:45









Chem_Student

32




32












  • There's no image. But in any case, you should use MathJax rather than an image.
    – Robert Israel
    Nov 29 at 3:57










  • I have updated the post. Also, there are many lines of working and I'm new to Maths Stack Exchange so I feel that in this moment in time neat, written working is the most appropriate way to go :)
    – Chem_Student
    Nov 29 at 4:00










  • The correct answer is $-115$.
    – Robert Israel
    Nov 29 at 4:01










  • See math.stackexchange.com/questions/3015661/…
    – lab bhattacharjee
    Nov 29 at 4:03


















  • There's no image. But in any case, you should use MathJax rather than an image.
    – Robert Israel
    Nov 29 at 3:57










  • I have updated the post. Also, there are many lines of working and I'm new to Maths Stack Exchange so I feel that in this moment in time neat, written working is the most appropriate way to go :)
    – Chem_Student
    Nov 29 at 4:00










  • The correct answer is $-115$.
    – Robert Israel
    Nov 29 at 4:01










  • See math.stackexchange.com/questions/3015661/…
    – lab bhattacharjee
    Nov 29 at 4:03
















There's no image. But in any case, you should use MathJax rather than an image.
– Robert Israel
Nov 29 at 3:57




There's no image. But in any case, you should use MathJax rather than an image.
– Robert Israel
Nov 29 at 3:57












I have updated the post. Also, there are many lines of working and I'm new to Maths Stack Exchange so I feel that in this moment in time neat, written working is the most appropriate way to go :)
– Chem_Student
Nov 29 at 4:00




I have updated the post. Also, there are many lines of working and I'm new to Maths Stack Exchange so I feel that in this moment in time neat, written working is the most appropriate way to go :)
– Chem_Student
Nov 29 at 4:00












The correct answer is $-115$.
– Robert Israel
Nov 29 at 4:01




The correct answer is $-115$.
– Robert Israel
Nov 29 at 4:01












See math.stackexchange.com/questions/3015661/…
– lab bhattacharjee
Nov 29 at 4:03




See math.stackexchange.com/questions/3015661/…
– lab bhattacharjee
Nov 29 at 4:03










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Your answer is correct. Here is an alternative approach.



You have
$$p^2+q^2+r^2=(underbrace{p+q+r}_{=-4})^2-2(underbrace{pq+pr+qr}_{=-4})=24,$$
By $x^3=-4x^2+4x-1$,
$$p^3+q^3+r^3=-4(p^2+q^2+r^2)+4(p+q+r)-(1+1+1)=-115.$$






share|cite|improve this answer





















  • Thank you for the quicker method, however I don't really understand if you used a formula for the first line, or if it's derived from elsewhere?
    – Chem_Student
    Nov 29 at 4:11










  • Expand $(p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr$.
    – Tianlalu
    Nov 29 at 4:13










  • Ok thank you so much!
    – Chem_Student
    Nov 29 at 4:15










  • Sorry, also, did you subsitute x=p+q+r, because I don't understand how $x^3=p^3+q^3+r^3$
    – Chem_Student
    Nov 29 at 4:20












  • Substitute $p,q,r$ separately, we get 3 equalities. Then add them together.
    – Tianlalu
    Nov 29 at 4:22


















up vote
0
down vote













Your solution is correct. You can also take a shortcut:
$$begin{align}p+q+r&=-4 Rightarrow \
p+q&=-4-r Rightarrow \
(p+q)^3&=(-4-r)^3 Rightarrow \
p^3+q^3+3pq(p+q)&=-64-48r-12r^2-r^3 Rightarrow \
p^3+q^3+r^3&=-3left(-frac1rright)(-4-r)-64-48r-12r^2=\
&=-12r^2-48r-67-frac{12}{r}=\
&=frac{-12(r^3+4r^2-4r+1)-115r}{r}=\
&=-115,
end{align}$$
because:
$$begin{align}p+q+r&=-4;\
pq+qr+rp&=-4;\
pqr&=-1;\
r^3+4r^2-4r+1&=0.end{align}$$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Your answer is correct. Here is an alternative approach.



    You have
    $$p^2+q^2+r^2=(underbrace{p+q+r}_{=-4})^2-2(underbrace{pq+pr+qr}_{=-4})=24,$$
    By $x^3=-4x^2+4x-1$,
    $$p^3+q^3+r^3=-4(p^2+q^2+r^2)+4(p+q+r)-(1+1+1)=-115.$$






    share|cite|improve this answer





















    • Thank you for the quicker method, however I don't really understand if you used a formula for the first line, or if it's derived from elsewhere?
      – Chem_Student
      Nov 29 at 4:11










    • Expand $(p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr$.
      – Tianlalu
      Nov 29 at 4:13










    • Ok thank you so much!
      – Chem_Student
      Nov 29 at 4:15










    • Sorry, also, did you subsitute x=p+q+r, because I don't understand how $x^3=p^3+q^3+r^3$
      – Chem_Student
      Nov 29 at 4:20












    • Substitute $p,q,r$ separately, we get 3 equalities. Then add them together.
      – Tianlalu
      Nov 29 at 4:22















    up vote
    1
    down vote



    accepted










    Your answer is correct. Here is an alternative approach.



    You have
    $$p^2+q^2+r^2=(underbrace{p+q+r}_{=-4})^2-2(underbrace{pq+pr+qr}_{=-4})=24,$$
    By $x^3=-4x^2+4x-1$,
    $$p^3+q^3+r^3=-4(p^2+q^2+r^2)+4(p+q+r)-(1+1+1)=-115.$$






    share|cite|improve this answer





















    • Thank you for the quicker method, however I don't really understand if you used a formula for the first line, or if it's derived from elsewhere?
      – Chem_Student
      Nov 29 at 4:11










    • Expand $(p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr$.
      – Tianlalu
      Nov 29 at 4:13










    • Ok thank you so much!
      – Chem_Student
      Nov 29 at 4:15










    • Sorry, also, did you subsitute x=p+q+r, because I don't understand how $x^3=p^3+q^3+r^3$
      – Chem_Student
      Nov 29 at 4:20












    • Substitute $p,q,r$ separately, we get 3 equalities. Then add them together.
      – Tianlalu
      Nov 29 at 4:22













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Your answer is correct. Here is an alternative approach.



    You have
    $$p^2+q^2+r^2=(underbrace{p+q+r}_{=-4})^2-2(underbrace{pq+pr+qr}_{=-4})=24,$$
    By $x^3=-4x^2+4x-1$,
    $$p^3+q^3+r^3=-4(p^2+q^2+r^2)+4(p+q+r)-(1+1+1)=-115.$$






    share|cite|improve this answer












    Your answer is correct. Here is an alternative approach.



    You have
    $$p^2+q^2+r^2=(underbrace{p+q+r}_{=-4})^2-2(underbrace{pq+pr+qr}_{=-4})=24,$$
    By $x^3=-4x^2+4x-1$,
    $$p^3+q^3+r^3=-4(p^2+q^2+r^2)+4(p+q+r)-(1+1+1)=-115.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 29 at 4:05









    Tianlalu

    3,01021038




    3,01021038












    • Thank you for the quicker method, however I don't really understand if you used a formula for the first line, or if it's derived from elsewhere?
      – Chem_Student
      Nov 29 at 4:11










    • Expand $(p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr$.
      – Tianlalu
      Nov 29 at 4:13










    • Ok thank you so much!
      – Chem_Student
      Nov 29 at 4:15










    • Sorry, also, did you subsitute x=p+q+r, because I don't understand how $x^3=p^3+q^3+r^3$
      – Chem_Student
      Nov 29 at 4:20












    • Substitute $p,q,r$ separately, we get 3 equalities. Then add them together.
      – Tianlalu
      Nov 29 at 4:22


















    • Thank you for the quicker method, however I don't really understand if you used a formula for the first line, or if it's derived from elsewhere?
      – Chem_Student
      Nov 29 at 4:11










    • Expand $(p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr$.
      – Tianlalu
      Nov 29 at 4:13










    • Ok thank you so much!
      – Chem_Student
      Nov 29 at 4:15










    • Sorry, also, did you subsitute x=p+q+r, because I don't understand how $x^3=p^3+q^3+r^3$
      – Chem_Student
      Nov 29 at 4:20












    • Substitute $p,q,r$ separately, we get 3 equalities. Then add them together.
      – Tianlalu
      Nov 29 at 4:22
















    Thank you for the quicker method, however I don't really understand if you used a formula for the first line, or if it's derived from elsewhere?
    – Chem_Student
    Nov 29 at 4:11




    Thank you for the quicker method, however I don't really understand if you used a formula for the first line, or if it's derived from elsewhere?
    – Chem_Student
    Nov 29 at 4:11












    Expand $(p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr$.
    – Tianlalu
    Nov 29 at 4:13




    Expand $(p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr$.
    – Tianlalu
    Nov 29 at 4:13












    Ok thank you so much!
    – Chem_Student
    Nov 29 at 4:15




    Ok thank you so much!
    – Chem_Student
    Nov 29 at 4:15












    Sorry, also, did you subsitute x=p+q+r, because I don't understand how $x^3=p^3+q^3+r^3$
    – Chem_Student
    Nov 29 at 4:20






    Sorry, also, did you subsitute x=p+q+r, because I don't understand how $x^3=p^3+q^3+r^3$
    – Chem_Student
    Nov 29 at 4:20














    Substitute $p,q,r$ separately, we get 3 equalities. Then add them together.
    – Tianlalu
    Nov 29 at 4:22




    Substitute $p,q,r$ separately, we get 3 equalities. Then add them together.
    – Tianlalu
    Nov 29 at 4:22










    up vote
    0
    down vote













    Your solution is correct. You can also take a shortcut:
    $$begin{align}p+q+r&=-4 Rightarrow \
    p+q&=-4-r Rightarrow \
    (p+q)^3&=(-4-r)^3 Rightarrow \
    p^3+q^3+3pq(p+q)&=-64-48r-12r^2-r^3 Rightarrow \
    p^3+q^3+r^3&=-3left(-frac1rright)(-4-r)-64-48r-12r^2=\
    &=-12r^2-48r-67-frac{12}{r}=\
    &=frac{-12(r^3+4r^2-4r+1)-115r}{r}=\
    &=-115,
    end{align}$$
    because:
    $$begin{align}p+q+r&=-4;\
    pq+qr+rp&=-4;\
    pqr&=-1;\
    r^3+4r^2-4r+1&=0.end{align}$$






    share|cite|improve this answer

























      up vote
      0
      down vote













      Your solution is correct. You can also take a shortcut:
      $$begin{align}p+q+r&=-4 Rightarrow \
      p+q&=-4-r Rightarrow \
      (p+q)^3&=(-4-r)^3 Rightarrow \
      p^3+q^3+3pq(p+q)&=-64-48r-12r^2-r^3 Rightarrow \
      p^3+q^3+r^3&=-3left(-frac1rright)(-4-r)-64-48r-12r^2=\
      &=-12r^2-48r-67-frac{12}{r}=\
      &=frac{-12(r^3+4r^2-4r+1)-115r}{r}=\
      &=-115,
      end{align}$$
      because:
      $$begin{align}p+q+r&=-4;\
      pq+qr+rp&=-4;\
      pqr&=-1;\
      r^3+4r^2-4r+1&=0.end{align}$$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Your solution is correct. You can also take a shortcut:
        $$begin{align}p+q+r&=-4 Rightarrow \
        p+q&=-4-r Rightarrow \
        (p+q)^3&=(-4-r)^3 Rightarrow \
        p^3+q^3+3pq(p+q)&=-64-48r-12r^2-r^3 Rightarrow \
        p^3+q^3+r^3&=-3left(-frac1rright)(-4-r)-64-48r-12r^2=\
        &=-12r^2-48r-67-frac{12}{r}=\
        &=frac{-12(r^3+4r^2-4r+1)-115r}{r}=\
        &=-115,
        end{align}$$
        because:
        $$begin{align}p+q+r&=-4;\
        pq+qr+rp&=-4;\
        pqr&=-1;\
        r^3+4r^2-4r+1&=0.end{align}$$






        share|cite|improve this answer












        Your solution is correct. You can also take a shortcut:
        $$begin{align}p+q+r&=-4 Rightarrow \
        p+q&=-4-r Rightarrow \
        (p+q)^3&=(-4-r)^3 Rightarrow \
        p^3+q^3+3pq(p+q)&=-64-48r-12r^2-r^3 Rightarrow \
        p^3+q^3+r^3&=-3left(-frac1rright)(-4-r)-64-48r-12r^2=\
        &=-12r^2-48r-67-frac{12}{r}=\
        &=frac{-12(r^3+4r^2-4r+1)-115r}{r}=\
        &=-115,
        end{align}$$
        because:
        $$begin{align}p+q+r&=-4;\
        pq+qr+rp&=-4;\
        pqr&=-1;\
        r^3+4r^2-4r+1&=0.end{align}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 at 6:57









        farruhota

        18.8k2736




        18.8k2736






























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            Guerrita

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            Guerrita Présentation Nom de naissance Rafael Guerra Bejarano Apodo Guerrita Naissance 6 mars 1862 Cordoue Décès 21 février 1941 (à 78 ans) Nationalité Espagnol Carrière Alternative 29 septembre 1887 à Madrid Parrain, Lagartijo Fin de carrière 15 octobre 1899 Distinction Nommé « Deuxième Calife de la Tauromachie » modifier   Rafael Guerra Bejarano dit « Guerrita » , né à Cordoue (Espagne) le 6 mars 1862, mort à Cordoue le 21 février 1941, était un matador espagnol. Présentation | « Guerrita » fut sans doute l’un des plus grands matadors de son époque, peut-être même le plus grand : insurpassable avec les banderilles, d’une parfaite maîtrise avec la muleta , infaillible avec l’épée. Il se fait d'abord banderillero comme il est fréquent à l'époque et acquiert vite une réputation qui dépasse celle du maestro qu’il assiste, d’abord « Bocanegra », puis Fernando Gómez « El Gall...
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