Mixing
Evaluating the value of the roots of the expression
$begingroup$
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
asked Jan 6 at 0:11
ZachZach
365
$endgroup$
add a comment |
$begingroup$
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
asked Jan 6 at 0:11
ZachZach
365
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 6 at 0:11
ZachZach
365
$endgroup$
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
asked Jan 6 at 0:11
ZachZach
365
asked Jan 6 at 0:11
ZachZach
365
edited Jan 6 at 0:13
Zach
asked Jan 6 at 0:11
ZachZach
365
asked Jan 6 at 0:11
ZachZach
365
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.
answered Jan 6 at 0:20
MindlackMindlack
3,11717
$endgroup$
add a comment |
$begingroup$
$$x^3+1=2x^2+3x$$
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 $a^3,b^3,c^3$
$$a^3+b^3+c^3=dfrac{23}1$$
answered Jan 6 at 10:15
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
By Vieta's formulas:
$$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
Then you can use the formulas:
$$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
&=2^2-2(-3)=\
&=10.end{align}$$
$$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
&=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
&=2(2^2-3(-3))+3(-1)=\
&=23.end{align}$$
$$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
&=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
&=10^2-2[(-3)^2-2(-1)2]=\
&=126.end{align}$$
$$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
& 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
&=2^5-frac{5[-15]}{3}(7)=\
&=207.end{align}$$
answered Jan 6 at 11:56
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
active
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$begingroup$
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, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.
answered Jan 6 at 0:20
MindlackMindlack
3,11717
$endgroup$
add a comment |
$begingroup$
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, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.
answered Jan 6 at 0:20
MindlackMindlack
3,11717
$endgroup$
add a comment |
$begingroup$
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, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.
answered Jan 6 at 0:20
MindlackMindlack
3,11717
$endgroup$
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, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.
answered Jan 6 at 0:20
MindlackMindlack
3,11717
answered Jan 6 at 0:20
MindlackMindlack
3,11717
answered Jan 6 at 0:20
MindlackMindlack
3,11717
answered Jan 6 at 0:20
MindlackMindlack
3,11717
3,11717
add a comment |
add a comment |
$begingroup$
$$x^3+1=2x^2+3x$$
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 $a^3,b^3,c^3$
$$a^3+b^3+c^3=dfrac{23}1$$
answered Jan 6 at 10:15
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
$$x^3+1=2x^2+3x$$
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 $a^3,b^3,c^3$
$$a^3+b^3+c^3=dfrac{23}1$$
answered Jan 6 at 10:15
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
$$x^3+1=2x^2+3x$$
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 $a^3,b^3,c^3$
$$a^3+b^3+c^3=dfrac{23}1$$
answered Jan 6 at 10:15
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$$x^3+1=2x^2+3x$$
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 $a^3,b^3,c^3$
$$a^3+b^3+c^3=dfrac{23}1$$
answered Jan 6 at 10:15
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 6 at 10:15
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 6 at 10:15
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 6 at 10:15
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
$begingroup$
By Vieta's formulas:
$$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
Then you can use the formulas:
$$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
&=2^2-2(-3)=\
&=10.end{align}$$
$$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
&=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
&=2(2^2-3(-3))+3(-1)=\
&=23.end{align}$$
$$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
&=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
&=10^2-2[(-3)^2-2(-1)2]=\
&=126.end{align}$$
$$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
& 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
&=2^5-frac{5[-15]}{3}(7)=\
&=207.end{align}$$
answered Jan 6 at 11:56
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
By Vieta's formulas:
$$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
Then you can use the formulas:
$$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
&=2^2-2(-3)=\
&=10.end{align}$$
$$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
&=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
&=2(2^2-3(-3))+3(-1)=\
&=23.end{align}$$
$$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
&=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
&=10^2-2[(-3)^2-2(-1)2]=\
&=126.end{align}$$
$$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
& 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
&=2^5-frac{5[-15]}{3}(7)=\
&=207.end{align}$$
answered Jan 6 at 11:56
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
By Vieta's formulas:
$$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
Then you can use the formulas:
$$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
&=2^2-2(-3)=\
&=10.end{align}$$
$$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
&=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
&=2(2^2-3(-3))+3(-1)=\
&=23.end{align}$$
$$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
&=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
&=10^2-2[(-3)^2-2(-1)2]=\
&=126.end{align}$$
$$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
& 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
&=2^5-frac{5[-15]}{3}(7)=\
&=207.end{align}$$
answered Jan 6 at 11:56
farruhotafarruhota
19.8k2738
$endgroup$
By Vieta's formulas:
$$a+b+c=2, ab+bc+ca=-3, abc=-1.$$
Then you can use the formulas:
$$begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\
&=2^2-2(-3)=\
&=10.end{align}$$
$$begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\
&=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\
&=2(2^2-3(-3))+3(-1)=\
&=23.end{align}$$
$$begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\
&=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\
&=10^2-2[(-3)^2-2(-1)2]=\
&=126.end{align}$$
$$begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\
& 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\
&=2^5-frac{5[2^3-B]}{3}(A+(-3))=\
&=2^5-frac{5[-15]}{3}(7)=\
&=207.end{align}$$
answered Jan 6 at 11:56
farruhotafarruhota
19.8k2738
answered Jan 6 at 11:56
farruhotafarruhota
19.8k2738
answered Jan 6 at 11:56
farruhotafarruhota
19.8k2738
answered Jan 6 at 11:56
farruhotafarruhota
19.8k2738
19.8k2738
add a comment |
add a comment |
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$begingroup$
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