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.
Expand the two equations:
1) abc - (ab + bc + ca) + (a + b +c) - 1 = abc -1 =>
(ab + bc + ca) = (a + b + c)
2) abc - 2(ab + bc + ca) + 4(a + b + c) - 8 = abc -2 =>
4(a + b + c) = 6 + 2(ab + bc + ca) =>
2(a + b + c) = 3 + (ab + bc + ca) => (using 1)
(a + b + c) = 3 (and so (ab+bc +ca) also =3)
if (a + b + c) = 3 then its square = 9:
(a^2 + b^2 + c^2) + 2(ab+bc+ca) = 9
(a^2 + b^2 + c^2) + 6 = 9
(a^2 + b^2 + c^2) = 3
now subtract (a+b+c) = 3 twice and add 3 to both sides:
(a^2 + b^2 + c^2 - 2(a + b + c) +3 = 3 -2*3 +3 = 0
rearranging,
(a-1)^2 + (b-1)^2 + (c-1)^2 = 0
Since the square of a real number is >= 0 and is equal only when the number itself is zero, this equation can only hold when all three expressions are zero:
(a-1)^2 = 0 => (a-1) = 0 => a = 1
ditto for b and c
So the only solution for (a,b,c) is (1,1,1)
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Posted by Paul
on 2023-09-23 16:13:07 |
Solution a different way
