An easy problem #2

Number Theory Level 3

If a+b+c=0 then the value of a 2 b c \frac {a^{2}}{bc} + b 2 a c \frac {b^{2}}{ac} + c 2 a b \frac {c^{2}}{ab} is


The answer is 3.

Vishal S by Vishal S , 21, India

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3 solutions

Vishal S
Jan 15, 2015

a 2 b c \frac {a^{2}}{bc} + b 2 a c \frac {b^{2}}{ac} + c 2 a b \frac {c^{2}}{ab}

⇒ \Rightarrow a 4 b c + b 4 a c + c 4 a b ( a b c ) 2 \frac {a^4bc+b^4ac+c^4ab}{(abc)^2}

⇒ \Rightarrow a b c ( a 3 + b 3 + c 3 ) ( a b c ) 2 \frac {abc(a^{3}+b^{3}+c^{3})}{(abc)^{2}}

Since a+b+c=0.Therefore a 3 a^{3} + b 3 b^{3} + c 3 c^{3} =3abc

⇒ \Rightarrow a b c ( a 3 + b 3 + c 3 ) ( a b c ) 2 \frac {abc(a^{3}+b^{3}+c^{3})}{(abc)^{2}} = a b c ( 3 a b c ) ( a b c ) 2 \frac {abc(3abc)}{(abc)^{2}}

Therefore a 2 b c \frac {a^{2}}{bc} + b 2 a c \frac {b^{2}}{ac} + c 2 a b \frac {c^{2}}{ab} when a+b+c=0 is 3 \boxed{3}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I get 0. a 2 b c + b 2 a c + c 2 a b = a 2 a c a b + b 2 b c a b + c 2 b c a c b c a c a b = a ( a c b ) + b ( b c a ) + c ( c b a ) b c a c a b = a b c ( a + b + c ) a 2 b 2 c 2 = a + b + c a b c S i n c e a + b + c = 0 ⟹ a + b + c a b c = 0 a b c = 0 \frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ac } +\frac { { c }^{ 2 } }{ ab } =\frac { { a }^{ 2 }acab+\begin{matrix} { b }^{ 2 }bcab+ & { c }^{ 2 } \end{matrix}bcac }{ bcacab } =\frac { a(acb)+b(bca)+c(cba) }{ bcacab } =\frac { abc(a+b+c) }{ \begin{matrix} { a }^{ 2 } & { b }^{ 2 } & { c }^{ 2 } \end{matrix} } =\frac { a+b+c }{ abc } \quad Since\quad a+b+c=0\Longrightarrow \frac { a+b+c }{ abc } =\frac { 0 }{ abc } =0

Thoma Zoto - 6 years, 3 months ago

Why does it equal 3abc?

Cédric Goemaere - 4 years, 7 months ago
Rama Devi
May 13, 2015

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Fox To-ong
Feb 9, 2015

that 's 1, -1 and 1

0 Helpful 0 Interesting 0 Brilliant 0 Confused

That doesnt give 0!

Cédric Goemaere - 4 years, 7 months ago

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