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NathanMasky
24 Feb 2025, 14:57
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
25% (02:15) correct
75%
(02:15)
wrong
based on 59
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Date
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What is the value of a + b + 1?
(1) \(a^3 + b^3 + 1 = 3ab\)
(2) \(a^2 - 2ab + b^2 > 0\)
(1) \(a^3 + b^3 + 1 = 3ab\)
(2) \(a^2 - 2ab + b^2 > 0\)
NathanMasky
25 Feb 2025, 09:52
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Statement I: a3 + b3 + 1 = 3ab
This equation is a bit tricky. We need to recognize a specific algebraic identity:
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - ac - bc)
If we let c = 1, we get:
a3 + b3 + 1 - 3ab = (a + b + 1)(a2 + b2 + 1 - ab - a - b)
We are given that a3 + b3 + 1 = 3ab. Therefore:
a3 + b3 + 1 - 3ab = 0
Substituting from the identity, we have:
(a + b + 1)(a2 + b2 + 1 - ab - a - b) = 0
This equation is satisfied if either:
a + b + 1 = 0 2. a2 + b2 + 1 - ab - a - b = 0
Let's examine the second case. Multiply both sides by 2:
2a2 + 2b2 + 2 - 2ab - 2a - 2b = 0
(a2 - 2ab + b2) + (a2 - 2a + 1) + (b2 - 2b + 1) = 0
(a - b)2 + (a - 1)2 + (b - 1)2 = 0
For the sum of squares to be zero, each term must be zero:
(a - b)2 = 0 => a = b
(a - 1)2 = 0 => a = 1
(b - 1)2 = 0 => b = 1
So, the second case leads to a = b = 1.
Conclusion for Statement I: We have two possibilities:
a + b + 1 = 0
a = 1 and b = 1 (which means a + b + 1 = 3)
Statement I alone is not sufficient.
Statement II: a2 - 2ab + b2 > 0
(a - b)2 > 0
This implies that a ≠ b.
Statement II alone is not sufficient.
Combining Statements I and II:
From Statement I, we know either a + b + 1 = 0 or a = b = 1.
From Statement II, we know a ≠ b.
Since a ≠ b, the case where a = b = 1 is eliminated.
Therefore, we must have a + b + 1 = 0.
Final Answer: The value of a + b + 1 is 0.
The correct answer is C.
This equation is a bit tricky. We need to recognize a specific algebraic identity:
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - ac - bc)
If we let c = 1, we get:
a3 + b3 + 1 - 3ab = (a + b + 1)(a2 + b2 + 1 - ab - a - b)
We are given that a3 + b3 + 1 = 3ab. Therefore:
a3 + b3 + 1 - 3ab = 0
Substituting from the identity, we have:
(a + b + 1)(a2 + b2 + 1 - ab - a - b) = 0
This equation is satisfied if either:
a + b + 1 = 0 2. a2 + b2 + 1 - ab - a - b = 0
Let's examine the second case. Multiply both sides by 2:
2a2 + 2b2 + 2 - 2ab - 2a - 2b = 0
(a2 - 2ab + b2) + (a2 - 2a + 1) + (b2 - 2b + 1) = 0
(a - b)2 + (a - 1)2 + (b - 1)2 = 0
For the sum of squares to be zero, each term must be zero:
(a - b)2 = 0 => a = b
(a - 1)2 = 0 => a = 1
(b - 1)2 = 0 => b = 1
So, the second case leads to a = b = 1.
Conclusion for Statement I: We have two possibilities:
a + b + 1 = 0
a = 1 and b = 1 (which means a + b + 1 = 3)
Statement I alone is not sufficient.
Statement II: a2 - 2ab + b2 > 0
(a - b)2 > 0
This implies that a ≠ b.
Statement II alone is not sufficient.
Combining Statements I and II:
From Statement I, we know either a + b + 1 = 0 or a = b = 1.
From Statement II, we know a ≠ b.
Since a ≠ b, the case where a = b = 1 is eliminated.
Therefore, we must have a + b + 1 = 0.
Final Answer: The value of a + b + 1 is 0.
The correct answer is C.
19 Apr 2026, 19:38
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