value of algebric expression
$begingroup$
If $a,b,c$ are three distinct complex number and $displaystyle frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}=0$ . Then $displaystyle frac{a^2}{(b-c)^2}+frac{b^2}{(c-a)^2}+frac{c^2}{(a-b)^2}=$
what i have try
$displaystyle bigg(frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}bigg)^2=sum_{cyc}frac{a^2}{(b-c)^2}+2frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$
$displaystyle sum_{cyc} frac{a^2}{(b-c)^2}=-2frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$
How do I find value of right side expression?
algebra-precalculus factoring
edited Dec 28 '18 at 6:20
Michael Rozenberg
108k1895200
asked Dec 28 '18 at 4:51
jackyjacky
1,211815
$endgroup$
add a comment |
$begingroup$
If $a,b,c$ are three distinct complex number and $displaystyle frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}=0$ . Then $displaystyle frac{a^2}{(b-c)^2}+frac{b^2}{(c-a)^2}+frac{c^2}{(a-b)^2}=$
what i have try
$displaystyle bigg(frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}bigg)^2=sum_{cyc}frac{a^2}{(b-c)^2}+2frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$
$displaystyle sum_{cyc} frac{a^2}{(b-c)^2}=-2frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$
How do I find value of right side expression?
algebra-precalculus factoring
edited Dec 28 '18 at 6:20
Michael Rozenberg
108k1895200
asked Dec 28 '18 at 4:51
jackyjacky
1,211815
$endgroup$
add a comment |
$begingroup$
If $a,b,c$ are three distinct complex number and $displaystyle frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}=0$ . Then $displaystyle frac{a^2}{(b-c)^2}+frac{b^2}{(c-a)^2}+frac{c^2}{(a-b)^2}=$
what i have try
$displaystyle bigg(frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}bigg)^2=sum_{cyc}frac{a^2}{(b-c)^2}+2frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$
$displaystyle sum_{cyc} frac{a^2}{(b-c)^2}=-2frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$
How do I find value of right side expression?
algebra-precalculus factoring
edited Dec 28 '18 at 6:20
Michael Rozenberg
108k1895200
asked Dec 28 '18 at 4:51
jackyjacky
1,211815
$endgroup$
If $a,b,c$ are three distinct complex number and $displaystyle frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}=0$ . Then $displaystyle frac{a^2}{(b-c)^2}+frac{b^2}{(c-a)^2}+frac{c^2}{(a-b)^2}=$
what i have try
$displaystyle bigg(frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}bigg)^2=sum_{cyc}frac{a^2}{(b-c)^2}+2frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$
$displaystyle sum_{cyc} frac{a^2}{(b-c)^2}=-2frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$
How do I find value of right side expression?
algebra-precalculus factoring
algebra-precalculus factoring
edited Dec 28 '18 at 6:20
Michael Rozenberg
108k1895200
asked Dec 28 '18 at 4:51
jackyjacky
1,211815
edited Dec 28 '18 at 6:20
Michael Rozenberg
108k1895200
asked Dec 28 '18 at 4:51
jackyjacky
1,211815
edited Dec 28 '18 at 6:20
Michael Rozenberg
108k1895200
edited Dec 28 '18 at 6:20
Michael Rozenberg
108k1895200
edited Dec 28 '18 at 6:20
Michael Rozenberg
108k1895200
108k1895200
asked Dec 28 '18 at 4:51
jackyjacky
1,211815
asked Dec 28 '18 at 4:51
jackyjacky
1,211815
asked Dec 28 '18 at 4:51
jackyjacky
1,211815
1,211815
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use the following factoring.
$$sum_{cyc}ab(a-b)=sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c).$$
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
You can check that the numerator and the denominator are equal but of different sign by simply distributing all the products. You can also work taking common factors in the numerator in several steps, until you reach the expression in the denominator. So, taking account of the minus sign affecting the whole expression, the answer would then be $1$.
Also, while the notation
$$sum frac{a^2}{(b-c)^2}$$
is an understandable shorthand in this context, it is of course mathematically wrong and even misleading if it got out of context.
answered Dec 28 '18 at 5:03
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
You can use the following factoring.
$$sum_{cyc}ab(a-b)=sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c).$$
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
You can use the following factoring.
$$sum_{cyc}ab(a-b)=sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c).$$
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
You can use the following factoring.
$$sum_{cyc}ab(a-b)=sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c).$$
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
You can use the following factoring.
$$sum_{cyc}ab(a-b)=sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c).$$
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
$begingroup$
You can check that the numerator and the denominator are equal but of different sign by simply distributing all the products. You can also work taking common factors in the numerator in several steps, until you reach the expression in the denominator. So, taking account of the minus sign affecting the whole expression, the answer would then be $1$.
Also, while the notation
$$sum frac{a^2}{(b-c)^2}$$
is an understandable shorthand in this context, it is of course mathematically wrong and even misleading if it got out of context.
answered Dec 28 '18 at 5:03
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
You can check that the numerator and the denominator are equal but of different sign by simply distributing all the products. You can also work taking common factors in the numerator in several steps, until you reach the expression in the denominator. So, taking account of the minus sign affecting the whole expression, the answer would then be $1$.
Also, while the notation
$$sum frac{a^2}{(b-c)^2}$$
is an understandable shorthand in this context, it is of course mathematically wrong and even misleading if it got out of context.
answered Dec 28 '18 at 5:03
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
You can check that the numerator and the denominator are equal but of different sign by simply distributing all the products. You can also work taking common factors in the numerator in several steps, until you reach the expression in the denominator. So, taking account of the minus sign affecting the whole expression, the answer would then be $1$.
Also, while the notation
$$sum frac{a^2}{(b-c)^2}$$
is an understandable shorthand in this context, it is of course mathematically wrong and even misleading if it got out of context.
answered Dec 28 '18 at 5:03
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
You can check that the numerator and the denominator are equal but of different sign by simply distributing all the products. You can also work taking common factors in the numerator in several steps, until you reach the expression in the denominator. So, taking account of the minus sign affecting the whole expression, the answer would then be $1$.
Also, while the notation
$$sum frac{a^2}{(b-c)^2}$$
is an understandable shorthand in this context, it is of course mathematically wrong and even misleading if it got out of context.
answered Dec 28 '18 at 5:03
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:03
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:03
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:03
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
add a comment |
add a comment |
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