Algebra

sanjoy18 wrote:If (ax+b)+3/(ax+b)=2 then value of [(ax+b)^3-(ax+b)]is

a)-2
b)-3
c)-6
d)-4
e)none of these

Are you sure you transcribed the question correctly?

Cheers,
Brent

sanjoy18 wrote:Hi Brent.

Actually i tried the question yesterday almost 2 hours but could not get correct result. so I posted in this forum..But Today morning I have tried again and get the solution..Here It is..
lets ax+b=p

then it is given that

p+3/p=2 then p^3-p???

now p+3/p=2
>>p^2=2p-3..............(A)
>> p^3=2p^2-3p
>>p^3=2(2p-3)-3p..using (A) again

>> p^3=4p-6-3p
>> p^3=p-6
>> p^3-p= -6
therefore Ans is -6

hence C

Let me know if you have issues to my approach

Regards,
Sanjoy

That's a very elaborate/tricky solution. Unless there's a different (nicer) solution here, this one might be past the 800+ range.

Now there is another solution, but this solution highlights the larger issue with this question. This question goes beyond the realm of real numbers, and GMAT questions stay within the realm of real numbers.

Here's what I mean:

As you noted, if we let ax + b = p, then we get the equation p + 3/p = 2
Multiply both sides by p to get: p² + 3 = 2p
Rearrange to get: p² - 2p + 3 = 0
Since we can't factor this quadratic equation, we can apply the quadratic formula.
When we do so, we get p = 1 + √(-2) or p = 1 - √(-2)
Since √(-2) is not a real number, p is not a real number, which means a, b and x are not real numbers.

Once we know that p = 1 + √(-2) or p = 1 - √(-2), we can plug these values into the expression p^3 - p and then evaluate.
Doing so, however, would require us to know how to deal with non-real (complex/imaginary) numbers, and this goes beyond the scope of the GMAT.

It's certainly a clever question, but I don't think it's something that anyone would see on test day.

Cheers,
Brent

cd86 wrote:
m^3-m => m( m^2-1) = (m-1)m (m+1)

now amongst the answer choices only -6 can be split into a multiplication of 3 consecutive numbers..so i chose -6

This approach is a great idea. However, it might not apply here.
(m-1), m, and (m+1) aren't 3 consecutive integers. In fact, they aren't even real numbers. As such, we can't make many conclusions about their product.

Cheers,
Brent