Problem 22 · 2021 Math Kangaroo
Stretch
Algebra & Patterns
substitution
The numbers a, b and c satisfy \(a+b+c=0\) and \(abc=78\). What is the value of \((a+b)(b+c)(c+a)\)?
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Answer: B — −78
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Hint 1 of 2
Since \(a+b+c=0\), each pair-sum equals the negative of the third number.
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Hint 2 of 2
Multiply the three rewritten factors together and use \(abc=78\).
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Approach: rewrite each factor using the zero sum
- From \(a+b+c=0\): \(a+b=-c\), \(b+c=-a\), and \(c+a=-b\).
- So \((a+b)(b+c)(c+a)=(-c)(-a)(-b)=-abc\).
- Given \(abc=78\), this is \(-78\).
- The value is −78, choice (B).
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