Grade 8 - Algebra #2

Algebra Level 3

Three complex numbers a , b , c a,~b,~c satisfy below three equations.

  • a + b + c = 2 , a+b+c=2,

  • a b + b c + c a = 1 , ab+bc+ca=1,

  • 1 a + 1 b + 1 c = 1 3 . \displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{3}.

Find the value of ( a − 1 ) ( b − 1 ) ( c − 1 ) (a-1)(b-1)(c-1) .

This problem is a part of <Grade 8 - Algebra> series .


The answer is 3.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Jul 8, 2017

1 a + 1 b + 1 c = a b + b c + c a a b c = 1 3 \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ca}{abc}=\dfrac{1}{3} .

∴ a b c = 3 \therefore~abc=3 .

( a − 1 ) ( b − 1 ) ( c − 1 ) = a b c − ( a b + b c + c a ) + ( a + b + c ) − 1 = 3 − 1 + 2 − 1 = 3 (a-1)(b-1)(c-1)=abc-(ab+bc+ca)+(a+b+c)-1=3-1+2-1=\boxed{3} .

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