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
105k1892198
asked Dec 28 '18 at 4:51
jackyjacky
847614
$endgroup$
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
105k1892198
asked Dec 28 '18 at 4:51
jackyjacky
847614
$endgroup$
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
105k1892198
asked Dec 28 '18 at 4:51
jackyjacky
847614
$endgroup$
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
105k1892198
asked Dec 28 '18 at 4:51
jackyjacky
847614
edited Dec 28 '18 at 6:20
Michael Rozenberg
105k1892198
asked Dec 28 '18 at 4:51
jackyjacky
847614
edited Dec 28 '18 at 6:20
Michael Rozenberg
105k1892198
edited Dec 28 '18 at 6:20
Michael Rozenberg
105k1892198
edited Dec 28 '18 at 6:20
Michael Rozenberg
105k1892198
105k1892198
asked Dec 28 '18 at 4:51
jackyjacky
847614
asked Dec 28 '18 at 4:51
jackyjacky
847614
asked Dec 28 '18 at 4:51
jackyjacky
847614
847614
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
105k1892198
$endgroup$
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
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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
105k1892198
$endgroup$
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
105k1892198
$endgroup$
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
105k1892198
$endgroup$
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
105k1892198
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
105k1892198
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
105k1892198
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
105k1892198
105k1892198
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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