Roots for a cubic equations
$begingroup$
Find all real or complex solutions of the simultaneous equations
$$begin{aligned}x+y+z&=3\x^2+y^2+z^2&=3\x^3+y^3+z^3&=end{aligned}$$
I assumed that $x,y,z$ are 3 roots for a cubic equation such that $$W^3+aW^2+bW+c=0$$
and using relationship between roots of a polynomial equation I got $a=-3,; b=3,; c=-1.$
Which means the normal cubic equation is $（W-1)^3.$
My question is: are there possible complex solutions for such the relationship between $x,y$ and $z$ still be true. And please help me find the complex solutions as well.
Thank you
polynomials complex-numbers
edited Dec 30 '18 at 20:54
user376343
3,9934829
asked Sep 27 '15 at 18:52
Joseph ZhuJoseph Zhu
112
$endgroup$
add a comment |
$begingroup$
Find all real or complex solutions of the simultaneous equations
$$begin{aligned}x+y+z&=3\x^2+y^2+z^2&=3\x^3+y^3+z^3&=end{aligned}$$
I assumed that $x,y,z$ are 3 roots for a cubic equation such that $$W^3+aW^2+bW+c=0$$
and using relationship between roots of a polynomial equation I got $a=-3,; b=3,; c=-1.$
Which means the normal cubic equation is $（W-1)^3.$
My question is: are there possible complex solutions for such the relationship between $x,y$ and $z$ still be true. And please help me find the complex solutions as well.
Thank you
polynomials complex-numbers
edited Dec 30 '18 at 20:54
user376343
3,9934829
asked Sep 27 '15 at 18:52
Joseph ZhuJoseph Zhu
112
$endgroup$
$begingroup$
what is with the third term?
$endgroup$
– Dr. Sonnhard Graubner
Sep 27 '15 at 18:56
2
$begingroup$
The cubic equation is incomplete.
$endgroup$
– Bernard
Sep 27 '15 at 18:57
add a comment |
$begingroup$
Find all real or complex solutions of the simultaneous equations
$$begin{aligned}x+y+z&=3\x^2+y^2+z^2&=3\x^3+y^3+z^3&=end{aligned}$$
I assumed that $x,y,z$ are 3 roots for a cubic equation such that $$W^3+aW^2+bW+c=0$$
and using relationship between roots of a polynomial equation I got $a=-3,; b=3,; c=-1.$
Which means the normal cubic equation is $（W-1)^3.$
My question is: are there possible complex solutions for such the relationship between $x,y$ and $z$ still be true. And please help me find the complex solutions as well.
Thank you
polynomials complex-numbers
edited Dec 30 '18 at 20:54
user376343
3,9934829
asked Sep 27 '15 at 18:52
Joseph ZhuJoseph Zhu
112
$endgroup$
Find all real or complex solutions of the simultaneous equations
$$begin{aligned}x+y+z&=3\x^2+y^2+z^2&=3\x^3+y^3+z^3&=end{aligned}$$
I assumed that $x,y,z$ are 3 roots for a cubic equation such that $$W^3+aW^2+bW+c=0$$
and using relationship between roots of a polynomial equation I got $a=-3,; b=3,; c=-1.$
Which means the normal cubic equation is $（W-1)^3.$
My question is: are there possible complex solutions for such the relationship between $x,y$ and $z$ still be true. And please help me find the complex solutions as well.
Thank you
polynomials complex-numbers
polynomials complex-numbers
edited Dec 30 '18 at 20:54
user376343
3,9934829
asked Sep 27 '15 at 18:52
Joseph ZhuJoseph Zhu
112
edited Dec 30 '18 at 20:54
user376343
3,9934829
asked Sep 27 '15 at 18:52
Joseph ZhuJoseph Zhu
112
edited Dec 30 '18 at 20:54
user376343
3,9934829
edited Dec 30 '18 at 20:54
user376343
3,9934829
edited Dec 30 '18 at 20:54
user376343
3,9934829
3,9934829
asked Sep 27 '15 at 18:52
Joseph ZhuJoseph Zhu
112
asked Sep 27 '15 at 18:52
Joseph ZhuJoseph Zhu
112
asked Sep 27 '15 at 18:52
Joseph ZhuJoseph Zhu
112
112
$begingroup$
what is with the third term?
$endgroup$
– Dr. Sonnhard Graubner
Sep 27 '15 at 18:56
2
$begingroup$
The cubic equation is incomplete.
$endgroup$
– Bernard
Sep 27 '15 at 18:57
add a comment |
$begingroup$
what is with the third term?
$endgroup$
– Dr. Sonnhard Graubner
Sep 27 '15 at 18:56
2
$begingroup$
The cubic equation is incomplete.
$endgroup$
– Bernard
Sep 27 '15 at 18:57
$begingroup$
what is with the third term?
$endgroup$
– Dr. Sonnhard Graubner
Sep 27 '15 at 18:56
$begingroup$
what is with the third term?
$endgroup$
– Dr. Sonnhard Graubner
Sep 27 '15 at 18:56
2
2
$begingroup$
The cubic equation is incomplete.
$endgroup$
– Bernard
Sep 27 '15 at 18:57
$begingroup$
The cubic equation is incomplete.
$endgroup$
– Bernard
Sep 27 '15 at 18:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.
answered Sep 27 '15 at 18:57
Dietrich BurdeDietrich Burde
82.8k649107
$endgroup$
add a comment |
$begingroup$
$textbf{Hint: }$ Try the method of Newton's Sums.$$$$
Let $$P_i=x_1^i+x_2^i+x_3^i$$
Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$
Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
Also, $$ e_1 = sum x_i = - frac{b}{a} $$
$$e_2 = sum x_i x_j = frac{ c } { a} $$
$$e_3 = sum x_i x_j x_k = - frac{ d } { a}
$$ $$$$
Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$
For $ i leq k - 1 $, we have the formulas
$$P_1 = e_1 times 1 $$
$$P_2 = e_1 P_1 - e_2 times 2 $$
$$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
$$$$
Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.
answered Jun 2 '18 at 7:50
IshanIshan
2,0221927
$endgroup$
add a comment |
$begingroup$
Cauchy Schwarz Inequality
$$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$
equality hold when $displaystyle x=y=z$
So we have $(x,y,z)=(1,1,1)$.
answered Jun 2 '18 at 8:04
DXTDXT
5,8872733
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.
answered Sep 27 '15 at 18:57
Dietrich BurdeDietrich Burde
82.8k649107
$endgroup$
add a comment |
$begingroup$
The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.
answered Sep 27 '15 at 18:57
Dietrich BurdeDietrich Burde
82.8k649107
$endgroup$
add a comment |
$begingroup$
The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.
answered Sep 27 '15 at 18:57
Dietrich BurdeDietrich Burde
82.8k649107
$endgroup$
The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.
answered Sep 27 '15 at 18:57
Dietrich BurdeDietrich Burde
82.8k649107
edited Sep 28 '15 at 17:48
answered Sep 27 '15 at 18:57
Dietrich BurdeDietrich Burde
82.8k649107
answered Sep 27 '15 at 18:57
Dietrich BurdeDietrich Burde
82.8k649107
answered Sep 27 '15 at 18:57
Dietrich BurdeDietrich Burde
82.8k649107
82.8k649107
add a comment |
add a comment |
$begingroup$
$textbf{Hint: }$ Try the method of Newton's Sums.$$$$
Let $$P_i=x_1^i+x_2^i+x_3^i$$
Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$
Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
Also, $$ e_1 = sum x_i = - frac{b}{a} $$
$$e_2 = sum x_i x_j = frac{ c } { a} $$
$$e_3 = sum x_i x_j x_k = - frac{ d } { a}
$$ $$$$
Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$
For $ i leq k - 1 $, we have the formulas
$$P_1 = e_1 times 1 $$
$$P_2 = e_1 P_1 - e_2 times 2 $$
$$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
$$$$
Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.
answered Jun 2 '18 at 7:50
IshanIshan
2,0221927
$endgroup$
add a comment |
$begingroup$
$textbf{Hint: }$ Try the method of Newton's Sums.$$$$
Let $$P_i=x_1^i+x_2^i+x_3^i$$
Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$
Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
Also, $$ e_1 = sum x_i = - frac{b}{a} $$
$$e_2 = sum x_i x_j = frac{ c } { a} $$
$$e_3 = sum x_i x_j x_k = - frac{ d } { a}
$$ $$$$
Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$
For $ i leq k - 1 $, we have the formulas
$$P_1 = e_1 times 1 $$
$$P_2 = e_1 P_1 - e_2 times 2 $$
$$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
$$$$
Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.
answered Jun 2 '18 at 7:50
IshanIshan
2,0221927
$endgroup$
add a comment |
$begingroup$
$textbf{Hint: }$ Try the method of Newton's Sums.$$$$
Let $$P_i=x_1^i+x_2^i+x_3^i$$
Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$
Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
Also, $$ e_1 = sum x_i = - frac{b}{a} $$
$$e_2 = sum x_i x_j = frac{ c } { a} $$
$$e_3 = sum x_i x_j x_k = - frac{ d } { a}
$$ $$$$
Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$
For $ i leq k - 1 $, we have the formulas
$$P_1 = e_1 times 1 $$
$$P_2 = e_1 P_1 - e_2 times 2 $$
$$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
$$$$
Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.
answered Jun 2 '18 at 7:50
IshanIshan
2,0221927
$endgroup$
$textbf{Hint: }$ Try the method of Newton's Sums.$$$$
Let $$P_i=x_1^i+x_2^i+x_3^i$$
Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$
Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
Also, $$ e_1 = sum x_i = - frac{b}{a} $$
$$e_2 = sum x_i x_j = frac{ c } { a} $$
$$e_3 = sum x_i x_j x_k = - frac{ d } { a}
$$ $$$$
Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$
For $ i leq k - 1 $, we have the formulas
$$P_1 = e_1 times 1 $$
$$P_2 = e_1 P_1 - e_2 times 2 $$
$$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
$$$$
Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.
answered Jun 2 '18 at 7:50
IshanIshan
2,0221927
edited Jun 2 '18 at 8:00
answered Jun 2 '18 at 7:50
IshanIshan
2,0221927
answered Jun 2 '18 at 7:50
IshanIshan
2,0221927
answered Jun 2 '18 at 7:50
IshanIshan
2,0221927
2,0221927
add a comment |
add a comment |
$begingroup$
Cauchy Schwarz Inequality
$$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$
equality hold when $displaystyle x=y=z$
So we have $(x,y,z)=(1,1,1)$.
answered Jun 2 '18 at 8:04
DXTDXT
5,8872733
$endgroup$
add a comment |
$begingroup$
Cauchy Schwarz Inequality
$$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$
equality hold when $displaystyle x=y=z$
So we have $(x,y,z)=(1,1,1)$.
answered Jun 2 '18 at 8:04
DXTDXT
5,8872733
$endgroup$
add a comment |
$begingroup$
Cauchy Schwarz Inequality
$$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$
equality hold when $displaystyle x=y=z$
So we have $(x,y,z)=(1,1,1)$.
answered Jun 2 '18 at 8:04
DXTDXT
5,8872733
$endgroup$
Cauchy Schwarz Inequality
$$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$
equality hold when $displaystyle x=y=z$
So we have $(x,y,z)=(1,1,1)$.
answered Jun 2 '18 at 8:04
DXTDXT
5,8872733
answered Jun 2 '18 at 8:04
DXTDXT
5,8872733
answered Jun 2 '18 at 8:04
DXTDXT
5,8872733
answered Jun 2 '18 at 8:04
DXTDXT
5,8872733
5,8872733
add a comment |
add a comment |
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$begingroup$
what is with the third term?
$endgroup$
– Dr. Sonnhard Graubner
Sep 27 '15 at 18:56
2
$begingroup$
The cubic equation is incomplete.
$endgroup$
– Bernard
Sep 27 '15 at 18:57