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

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

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

ab=-6
consider b=4, then a=-3/2 which is not a integer. But according to the question a,b are non zero integer. So, answer is A

I just have this question, why cant be a=1 or -2. Why cant the denominator equal zero..

Division by 0 is not allowed: anything/0 is undefined.

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?

\(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\)

Take 3ab common in the numerator.

\(\frac{3(ab) * [(ab)^2 + 3(ab) - 18]}{(a-1)(a + 2)} = 0\)

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 \((ab)^2 + 3ab - 18 = (ab + 6)*(ab - 3)\)


\(\frac{3(ab) * [(ab + 6)*(ab - 3)]}{(a-1)(a + 2)} = 0\)

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.

Answer (A)

Why does the 3ab which is factored out not translate to ab=0 as a possible answer?

The prompt indicates that a and b are NONZERO INTEGERS.
Thus, ab=0 is not a valid case.

Please tell me where i am going wrong.Bunuel VeritasKarishma chetan2u

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.

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