Substitute? Or not?

Algebra Level 4

Let a , b , c a, b, c be rational numbers such that c = − a b a + b c=-\frac { ab }{ a+b } . Which of the following expressions is correct?

This problem is part of the set Hard Equations

( a + b + 2 c ) 2 = a 2 + b 2 − 4 c 2 { (a+b+2c) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }-4{ c }^{ 2 } ( a + b + c ) 3 = a 3 + b 3 + c 3 { (a+b+c) }^{ 3 }={ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } ( a + b + c ) 2 = a 2 + b 2 + c 2 { (a+b+c) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } ( a + b + c ) ( a 2 + b 2 + c 2 ) = a 3 + b 3 + c 3 + a b c (a+b+c)({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })={ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }+abc a + b + c = a 3 + b 3 + c 3 a+b+c={ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }

Julian Uy by Julian Uy , 26, Philippines

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1 solution

Anandhu Raj
Feb 2, 2015

Given c = − a b a + b c=-\frac { ab }{ a+b }

We could rewrite as c ( a + b ) = − a b c(a+b)=-ab

⟹ c a + c b = − a b \Longrightarrow ca+cb=-ab

⟹ a b + b c + a c = 0 \Longrightarrow ab+bc+ac=0 ------------------(1)

Also we know that ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( a b + b c + a c ) { (a+b+c) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2(ab+bc+ac)

Using ( 1 ) (1) ,

( a + b + c ) 2 = a 2 + b 2 + c 2 \boxed{{ (a+b+c) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

good solution

Sai Ram - 5 years, 10 months ago

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Thank you :)

Anandhu Raj - 5 years, 10 months ago

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I up voted it

Sai Ram - 5 years, 10 months ago

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