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we cannot have a=1, and a=-2. because this will make denominator zero.

now, consider the numerator. let ab=k thus we have k(3k^2 +9k-54) =k(3k^2+18k-9k-54) =k(3k(k+6)-9(k+6)) =k(3k-9)(k+6)

if 3k-9=0 3k=9 k=3 i.e. ab=3 now if b=3, then a=1, which is not acceptable as a=1, will make the denominator equal to zero. if k+6=0 k=-6 or ab=-6

again, if b=3, then a=-2, which is not acceptable as a=-2, will make the denominator equal to zero.

thus except b=3, all other values could be possible. hence answer should be D

As ab=-6 or ab=3, so if we consider value of b as 4, a becomes Fraction in both the cases, but a is a non-zero integer(Given in the question). Hence b cannot be 4, so the answer should be (A) I only

Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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06 Nov 2014, 11:56

2

This post received KUDOS

IMO ans shud be A

using numerator and taking ab= x we get 3x(x^2+3x-18)=0 gives 2 equations 3x=0 .i.e 3 ab =0 i.e a=0 or b=0 # but questions has given both are non zero so it can be true therefore x^2-3x+18=0 , gives can be factorized as (x-6)(x+3)=0 gives x=6 or ab=6 it has soloutions as (1,6),(6,1),(2,3),3,2) a=1 and a=2 is not possible as questions will become non mathematical. Therfore solutions possible are (6,1) and (3,2) Therfore ans is A b=2

if b = 3, then a = -2 (Does not satisfy denominator)

For ab = 3

if b = 3, then a = 1 (Does not satisfy denominator)

Answer = A

One question here.

For For ab = 3, why cant we have: ab=3 --> ab=1(3), cannot be ab=3 --> ab=3(1), couldn't this one also be? Or it is not included in the solution because there is not "1" as an answer option?

Note that \((ab)^2 + 3ab - 18\) is a quadratic which we can split into factors just like we do for \(x^2 + 3x - 18\). \(x^2 + 3x - 18 = (x + 6)(x - 3)\) so \((ab)^2 + 3ab - 18 = (ab + 6)*(ab - 3)\)

Now, when will this fraction be 0? When one of the factors in the numerator is 0. (Note that denominator cannot be 0. So a cannot the 1 or -2.)

So either ab = 0 (a and b are non zero so not possible) or (ab + 6) = 0 i.e. ab = -6 or (ab - 3) = 0 i.e. ab = 3

So, we get that one of these two must hold. Either ab = -6 or ab = 3.

Now let's look at the possible values of b. Can b be 2? If b = 2, a = -3 (in this case, ab = -6. Satisfies) Can b be 3? If b = 3, a is -2 or 1. Both values are not possible so b cannot be 3. Can b be 4? If b = 4, a is not an integer in either case. So b cannot be 4.

Can you please provide a proper solution to this question? I am not very clear from the solutions provided above.

Thanks.

HI,

If someone has a problem in factoring an euatio, we can apply logic and substitute to get to our answer..

Quote:

If \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\), and a and b are both non-zero integers, which of the following could be the value of b?

I. 2 II. 3 III. 4

(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III

Now \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\) to be TRUE, \(3(ab)^3 + 9(ab)^2 - 54ab\) or \(3ab((ab)^2 + 9(ab) - 54\) should be 0 but with following restrictions... a)\(a\neq{0}\)...........................\(b\neq{0}\)...... It is given that a and b are non-zero b) \(a\neq{1}\)...........................\(a\neq{-2}\).........WHY? because the denominator will become 0 and the equation undefined..

so first substitute value of b given in choices abd check if substituting a as 1 or -2 gives us value of that equation as 0..

I. 2 \((ab)^2 + 3(ab) - 18\) ..................\((a*2)^2 + 3(a*2) - 18.....................4a^2+6a-18.................\) when a= 1, ans is 4+6-18 = -8 OK when a =-2, ans is 4*4-6*2-18=16-12-18 = 14..OK

so 2 is a valid value..

II. 3 \((ab)^2 + 3(ab) - 18\) ..................\((a*3)^2 + 3(a*3) - 18.....................9a^2+9a-18.................\) when a= 1, ans is 9+9-18 = 0 NO

so 3 is NOT a valid value..

III. 4 \((ab)^2 + 3(ab) - 18\) ..................\((a*4)^2 + 3(a*4) - 18.....................16a^2+12a-18.................\) when a= 1, ans is 16+12-18 = 10 OK when a =-2, ans is 16*4-12*2-18=64-24-18 = 22..OK

Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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12 Jul 2017, 19:01

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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24 Jul 2017, 09:40

Given (a-1) and (a+2) in the denominator, a is not equal to 1 or -2 Simplifying the numerator we get: [3(ab)^3 + 9(ab)^2 – 54ab] = 3ab[(ab)^2+3ab-18] = 3ab[(ab)^2+6ab-3ab-18] = 3ab(ab+6)(ab-3) Since the whole equation is equal to 0, we can say the numerator is also equal to 0. Therefore, (ignoring 3ab=0) ab+6=0 which implies b = -6/a -------say equ(1) ab-3=0 which implies b = 3/a --------say equ(2) Since a and b are both non-zero integers (note, they could be negative) from equ (2) we can say a = {-3,-1,1,3} However, applying the constraint set by the denominator, a = {-3,-1,3} Substituting these values of 'a' in the equ(1) and equ(2), we get b = {-3,-2,-1,1,2,6} Hence, the answer is option A.

Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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07 Aug 2017, 18:37

Start with initial analysis of the problem:

a and b are integers ab cannot = 0 We cannot have a = 1 or a = -2, since denominator cannot be 0 Since we want the solution = 0 , then the numerator must = 0

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