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Mathematics 16 Online
OpenStudy (anonymous):

ok could some one help me please

OpenStudy (anonymous):

\[\frac{ c ^{2}ab }{ \left( c-a \right)\left( b-c \right) }-\frac{ b ^{2}ca }{ \left( b-a \right)\left( b-c \right) }-\frac{ a ^{2}bc }{ \left( a-b \right)\left( a-c \right) }\]

OpenStudy (anonymous):

step 1 substitute (a-b)(a-c)= (b-a) (c-a) [multiply and check for verification ) step 2 now with the denominator terms being (c-a) ,(b-a), (b-c) ... multiply each numerator with the term missing in their respective denominator step 3 now u get the common denominator (b-a)(b-c)(c-a)... in the numerator take 'abc' commonly outside and expand.. u ll c all the terms getting cancelled.. answer abc/(b-a)(b-c)(c-a)

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