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a^144-1/(a^72+1)
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05 Jan 2020, 01:47
05 Jan 2020, 01:47
Expert Reply
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Question Stats:
73% (01:01) correct
26% (01:51) wrong based on 34 sessions
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Quantity A - \(\frac{a^{144}-1}{(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
Quantity B - \(a^{18}+1 \)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity B - \(a^{18}+1 \)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Re: a^144-1/(a^72+1)
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05 Jan 2020, 04:05
05 Jan 2020, 04:05
1
Carcass wrote:
Quantity A - \(\frac{a^{144}-1}{(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
Quantity B - \(a^{18}+1 \)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity B - \(a^{18}+1 \)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
QTY A:: \(\frac{a^{144}-1}{(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
= \(\frac{(a^{72})^2 -{1^2}} {(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
= \(\frac {(a^{72} +1)(a^{72}-1)} {(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
= \(\frac {(a^{72} +1)((a^{36})^2-(1^2))} {(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
= \(\frac {(a^{72} +1)(a^{36}+1) (a^{36}-1)} {(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
= \(\frac {(a^{72} +1)(a^{36}+1) ((a^{18})^2-(1^2))} {(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
= \(\frac {(a^{72} +1)(a^{36}+1) (a^{18} +1)(a^{18} - 1)} {(a^{72}+1)(a^{36} + 1)(a^{18}+1)}\)
= \((a^{18} - 1)\)
QTY B :: \((a^{18} + 1)\)
Therefore QTY A < QTY B
