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
104k1891196
asked Dec 28 '18 at 4:51
jackyjacky
866612
$endgroup$
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
104k1891196
asked Dec 28 '18 at 4:51
jackyjacky
866612
$endgroup$
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
104k1891196
asked Dec 28 '18 at 4:51
jackyjacky
866612
$endgroup$
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
104k1891196
asked Dec 28 '18 at 4:51
jackyjacky
866612
edited Dec 28 '18 at 6:20
Michael Rozenberg
104k1891196
asked Dec 28 '18 at 4:51
jackyjacky
866612
edited Dec 28 '18 at 6:20
Michael Rozenberg
104k1891196
edited Dec 28 '18 at 6:20
Michael Rozenberg
104k1891196
edited Dec 28 '18 at 6:20
Michael Rozenberg
104k1891196
104k1891196
asked Dec 28 '18 at 4:51
jackyjacky
866612
asked Dec 28 '18 at 4:51
jackyjacky
866612
asked Dec 28 '18 at 4:51
jackyjacky
866612
866612
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
104k1891196
$endgroup$
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
104k1891196
$endgroup$
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
104k1891196
$endgroup$
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
104k1891196
$endgroup$
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
104k1891196
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
104k1891196
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
104k1891196
answered Dec 28 '18 at 5:41
Michael RozenbergMichael Rozenberg
104k1891196
104k1891196
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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