The three musketeers!

Algebra Level 2

Consider 3 real numbers a , b , c 0 a,b,c \neq 0 satisfying a + b + c = 0 a+b+c=0

Find the value a 2 b c + b 2 a c + c 2 a b \dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}


This is an original problem and belongs to the set My creations

6 2 3 9

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2 solutions

Vilakshan Gupta
Sep 18, 2017

a 2 b c + b 2 a c + c 2 a b = a 3 + b 3 + c 3 a b c a 3 + b 3 + c 3 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 a b b c c a ) a 3 + b 3 + c 3 = 3 a b c a 3 + b 3 + c 3 a b c = 3 a b c a b c = 3 \dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc} \\ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\ \implies a^3+b^3+c^3=3abc \\ \therefore \frac{a^3+b^3+c^3}{abc}=\frac{3{abc}}{{abc}}=\boxed{3}

a + b +c =0 a + b = -c now raise both sides to power 3 you will get a^3 + b^3 + 3ab(a+b) = -c^3 in next step a^3 + b^3 + 3ab(-c) = -c^3 a^3 + b^3 +c^3 =3abc

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