Find the only possible triplet (a,b,c) of real numbers that satisfy this system of equations
- (a-1)(b-1)(c-1) = a*b*c - 1
- (a-2)(b-2)(c-2) = a*b*c - 2
Prove that no further triplet satisfy the given conditions.
**** Computer program/ excel solver assisted solutions are welcome, but a semi-analytic - namely, p&p and hand calculator based methodology is preferred.
I'll start with multiplying each equation out, canceling out like terms, and then isolating ab+ac+bc in each equation. That leaves us with:
ab+ac+bc = a+b+c
ab+ac+bc = 2(a+b+c)-3
Then a+b+c = 2(a+b+c)-3, which means a+b+c=3, and then ab+bc+ac=3.
Now define f(x) as the cubic polynomial which has a,b,c as its roots.
Then f(x) = (x-a)*(x-b)*(x-c) = x^3 - (a+b+c)x^2 + (ab+bc+ac)x - abc. Now substitute to get f(x) = x^3 - 3x^2 + 3x - abc.
Then f(x) = (x-1)^3 - (abc-1). Due to how f was constructed, it has three real roots a, b, and c. Then (x+1)^3 - (abc-1) = 0 must have three real roots.
But then solving the last equation yields x = -1+cbrt[abc-1], where cbrt is the multivalued cubic root with three branches. Two of these branches will be complex, except in the case that the value of abc-1 is zero. We seek three real roots, so it must be the case that abc-1=0.
Then abc=1, and then f(x) = (x-1)^3, which makes a=b=c=1 as the only solution where all values are real numbers. (a,b,c)=(1,1,1) is the unique real valued solution.
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