In step 4, cancelling off is the trick since cancelling off requires division on the two sides by a non zero number, and in the example a+b-c = 0 since a+b=c. Division by 0, gives indeterminable values and "leads to" the crazy "result".

I concur with Brent - A great magic trick.

Another one!
1*0 = 0*0
1 = 0 (Cancelling 0's from both sides)
Is it?

yes, since a+b-c =0 , So, actually it is 4*0=3*0

lol - here is another one

(-2)^2 = (2)^2

taking root on both sides:

-2 = 2 !!!!!

0 = 4 !!!!

You were to come up with some sample usage for this trick Aneesh ...

Hi,

Thanks for reminding me about it.
Brent, Anil and others: Thanks for resisting the temptation of giving away the trick.
Aanderson: Your trick was quite amusing, to say the least.
eagleeye, hey_thr67: you were absolutely correct. We cannot cancel off 0s from LHS and RHS.

Ok now, let's get down to business.

Let's start with a few basic DS problems.

1] What is the value of x?

(1) x^2 = 2x
(2) x is an even integer

2] Is |x| > 1?

(1) x^3 > x
(2) |x| = x

1] What is the value of x?

(1) x^2 = 2x

x*(x-2) = 0
x = 0 or x = 2. Insufficient!

(2) x is an even integer

x can be any even integer. Insufficient!

From 1 and 2

x can be 0 or 2. Insufficient

IMO E

2] Is |x| > 1?
The question can be rephrased to Is x>1 or x<-1?

(1) x^3 > x

x^3 - x > 0.
x*(x^2 -1) > 0.
x*(x-1)*(x+1)>0.
Let us now check the values which satisfy the inequation.
If x<-1, then x*(x-1)*(x+1)<0.
If -1<x<0, then x*(x-1)*(x+1)>0.
If 0<x<1, then x*(x-1)*(x+1)<0.
If x>1 then x*(x-1)*(x+1)>0.
So, If x*(x-1)*(x+1)>0, then -1<x<0 or x>1. Since x can be any number between -1 and 0 or any number greater than 1, statement 1 is insufficient to answer the question.

(2) |x| = x

Implies x>0
Since x can be any value between 0 and 1 and any number greater than 1, statement 2 is insufficient to answer the question.

1+2

Intersection of -1<x<0 or x>1 and x>0 is x>1. So 1 + 2 combined is sufficient to answer the question.
IMO C