Another huge problem!

Algebra Level 4

Simplify: a 3 ( a − b ) ( a − c ) ( a − d ) ( a − e ) + b 3 ( b − a ) ( b − c ) ( b − d ) ( b − e ) + … + e 3 ( e − a ) ( e − b ) ( e − c ) ( e − d ) . \dfrac {a^3}{(a-b)(a-c)(a-d)(a-e)}+\dfrac {b^3}{(b-a)(b-c)(b-d)(b-e)}+\ldots+\dfrac {e^3}{(e-a)(e-b)(e-c)(e-d)}.

5 a b + b c + … + e a \dfrac {a}{b}+\dfrac {b}{c}+\ldots+\dfrac {e}{a} 1 a b c d e abcde 6 0

Rohit Udaiwal by Rohit Udaiwal , 20, India

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2 solutions

Divyanshu Mehta
Dec 26, 2015

See in the above question one can notice that there will be a - sign after the odd numbered terms so basically therms are ought to get cancelled and on the options listed the only such ans is 0

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Please post a proper solution

Wen Z - 4 years, 9 months ago
Ritik Agrawal
Aug 21, 2017

If we cross multiple and observe the numerator we find a b c d e are roots of it... Each term has a max power 4...since there are 5 roots the expression must be an identity hence 0.

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