A fun algebra question

Algebra Level 3

a , b , a, b, and c c are real numbers such that a + x 2 = 2015 a+x^{2} = 2015 , b + x 2 = 2016 b+x^{2}= 2016 , c + x 2 = 2017 c+x^{2} = 2017 and a b c = 999 a b c= 999 . Find a b c + c a b + b a c 1 a 1 b 1 c \frac{a}{bc}+\frac{c}{ab}+\frac{b}{ac}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}

1 1 1 666 \frac{1}{666} 2016 2016 1 333 \frac{1}{333} 0 0

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1 solution

Rishabh Jain
Jun 16, 2016

Substract three eqns pairwise to get: a b = b c = 1 , a c = 2 a-b=b-c=-1,a-c=-2 .

Now write the given expression as:

a 2 + b 2 + c 2 a b b c c a a b c \dfrac{a^2+b^2+c^2-ab-bc-ca}{abc}

= ( a b ) 2 + ( b c ) 2 + ( c a ) 2 2 a b c =\dfrac{(a-b)^2+(b-c)^2+(c-a)^2}{2abc} = 1 + 1 + 4 2 ( 999 ) =\dfrac{1+1+4}{2(999)}

= 1 333 \Large =\boxed{\dfrac1{333}}

Aww man you beat me to it XD

Manuel Kahayon - 4 years, 12 months ago

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This was simple And I'm 13

Owen Moriarty - 4 years, 12 months ago

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