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.Please tell me where i am going wrong.
Bunuel VeritasKarishma chetan2u3(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.
So in such case making quadratic is better option to get all values .I often make such mistakes.
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