What is the value of a^4 - b^4?
1) a^2 - b^2 = 16
2) a + b = 8
Sorry there is no OA. The problems were handed out in class.
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Frankenstein
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Hi,
a^4 - b^4 = (a^2+b^2)(a^2-b^2)
From(1): a^2-b^2=16. We don't know the value of a^2+b^2
Insufficient
From(2):a+b=8
Insufficient
Both(1)&(2): a^2-b^2=(a+b)(a-b)=16 =>8(a-b)=16 => a-b=2. So a,b are known and a^4-b^4 can be calculated.
Hence C
a^4 - b^4 = (a^2+b^2)(a^2-b^2)
From(1): a^2-b^2=16. We don't know the value of a^2+b^2
Insufficient
From(2):a+b=8
Insufficient
Both(1)&(2): a^2-b^2=(a+b)(a-b)=16 =>8(a-b)=16 => a-b=2. So a,b are known and a^4-b^4 can be calculated.
Hence C
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cans
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a^4-b^4 = (a^2-b^2)(a^2+b^2)
a) insufficient as only (a^2-b^2)=16 is given
b)a+b=8. We are not given what is (a-b) thus insufficient,
a&b) (a-b)(a+b)=16 and (a+b)=8
thus a-b = 2
solving for a,b -> a=5,b=3
thus a^4-b^4 = 5^4-3^4 = 625-81 = 744
IMO C
a) insufficient as only (a^2-b^2)=16 is given
b)a+b=8. We are not given what is (a-b) thus insufficient,
a&b) (a-b)(a+b)=16 and (a+b)=8
thus a-b = 2
solving for a,b -> a=5,b=3
thus a^4-b^4 = 5^4-3^4 = 625-81 = 744
IMO C
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Cans!!
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aftableo2006
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