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b v/s 3

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KarunMendiratta
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b v/s 3 [#permalink] New post   16 Aug 2021, 08:27
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D

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3(ab)3+9(ab)2−54ab(a−1)(a+2)=0, where a and b are both non-zero integers.

Quantity A
Quantity B
b
3


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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D
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Re: b v/s 3 [#permalink] New post   16 Aug 2021, 09:53
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a≠1, a≠-2
(ab)2+3(ab)−18=0
substitute ab with u
u2+3u−18=0, (u+6)(u−3)=0 and ab=-6 or ab=3
only, when a=1, b is 3 and a≠1
only, when a=-2, b is 3 and a≠-2. Clearly, b cannot be 3 and excluding choice C.
where a and b are both non-zero integers, we conclude the possible sets for ab=-6 condition such as a={-6,-3,-1,2,3,6}
and b={1,2,6 ....} don't even need to continue since the answer must be D


KarunMendiratta wrote:
3(ab)3+9(ab)2−54ab(a−1)(a+2)=0, where a and b are both non-zero integers.

Quantity A
Quantity B
b
3


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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Re: b v/s 3 [#permalink] New post   27 Sep 2022, 22:05
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Re: b v/s 3 [#permalink]
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