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

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Manager
Joined: 17 Jul 2017
Posts: 119
If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

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17 Nov 2018, 21:25
chetan2u wrote:
vanam52923 wrote:
Bunuel wrote:

Tough and Tricky questions: Algebra.

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

Kudos for a correct solution.

3(ab)^3 + 9 (ab)^2=54 ab
3(ab)^2[ab+3]=54 ab
div by 3 ab b/s
ab[ab+3]=18
let ab =z
z[z+3]=18
z[z+3]=3*6
hence, ab=3
so a cant b 1 so b cnt b 3
but m not getting rest of soltuins?i know i some how messed up by not making quadratic equation ,can u please guide me where i m going wrong?

You are missing out on other value of z...
z(z+3)=18..
Here z can be -6 too.
-6(-6+3)=-6*-3=18
So ab=-6..
when a is -3, b is 2.... So I is valid..

Now for b as 4, not possible, otherwise a will become fraction.

chetan2u Thank you so much sir,I am big idiot.:P
So in such case making quadratic is better option to get all values .I often make such mistakes.
Manager
Joined: 23 Jul 2015
Posts: 64
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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19 Apr 2019, 20:47
I finally got the solution, but does anyone know how to solve this under 2 mins?
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo   [#permalink] 19 Apr 2019, 20:47

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