AlgebraDifficulty 5.1AIME, harderProve itUnited States
Problem: Let a, b, c, x, y, and z be complex numbers such that a=x−2b+c,b=y−2c+a,c=z−2a+b. If xy+yz+zx=67 and x+y+z=2010, find the value of xyz.
Solution
Solution: Manipulate the equations to get a common denominator: a=x−2b+c⟹x−2=ab+c⟹x−1=aa+b+c⟹x−11=a+b+ca; similarly, y−11=a+b+cb and z−11=a+b+cc. Thus x−11+y−11+z−11(y−1)(z−1)+(x−1)(z−1)+(x−1)(y−1)xy+yz+zx−2(x+y+z)+3xyz−2(xy+yz+zx)+3(x+y+z)−4xyz−2(67)+3(2010)−4xyz=1=(x−1)(y−1)(z−1)=xyz−(xy+yz+zx)+(x+y+z)−1=0=0=−5892
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