When You See a(b + c)/a, Cancel First — JEE Almost Always Means It

JEE Advanced questions are not random. Every ugly expression has been designed to collapse cleanly if you read it the right way. When you notice the same factor sitting in the numerator and denominator — \tfrac{a(b+c)}{a}, \tfrac{\sin x \cdot (x^2 - 4)}{\sin x}, \tfrac{(x-3)(x+7)}{x-3} — the setter is waving a flag at you. Cancel first. Expand second. Always.

The move in one line

\frac{a(b + c)}{a} \;=\; b + c, \qquad (a \neq 0).

One stroke. The a on top and the a on bottom divide to give 1, and multiplying by 1 does nothing. What is left is exactly the bracketed sum.

The move relies on a single rule — multiplying and dividing by the same non-zero quantity leaves the expression unchanged — applied to a common factor. Nothing more subtle is happening. What is subtle is recognising that the common factor is there.

Why the recognition is hard

Most students see \tfrac{a(b+c)}{a} and reflexively want to multiply out: ab + ac on top, divided by a, equals b + c. Same answer, more work. On a three-hour paper, that extra work eats the minutes you need for the last two questions.

The recognition failure is almost always a reading failure. Your eye lands on "numerator" and "denominator" as two separate objects, and you start simplifying each in isolation. The fix is to read the fraction as a ratio first and look for shared factors before doing anything else. Why this is faster: cancellation is one step (delete the factor on both sides). Expansion is two steps (multiply out the top, then divide each term). And expansion introduces fresh terms that you might misalign. Fewer steps, fewer mistakes.

A JEE-style example

From a mock paper:

\frac{\sin x \cdot (x^2 - \pi^2)}{\sin x \cdot (x - \pi)} \quad \text{for } x \neq \pi, \; x \neq n\pi.

Ignore the limit wrappers for a second — just simplify.

Wrong instinct. Expand \sin x \cdot (x^2 - \pi^2) in the numerator, then struggle to divide by the denominator term by term.
Right instinct. Spot two common factors: \sin x on top and bottom, and (x - \pi) hidden inside x^2 - \pi^2 = (x-\pi)(x+\pi) on top.

Cancel both:

\frac{\sin x \cdot (x-\pi)(x+\pi)}{\sin x \cdot (x - \pi)} \;=\; x + \pi.

The whole expression is just x + \pi. If a limit then asks for x \to \pi, the answer is 2\pi — immediate. Students who expand first do not get there in time.

The recognition pattern

Every time you see a fraction, run three quick checks before you do any arithmetic.

Is the same literal factor on top and bottom? \sin x, e^x, a variable, a polynomial in brackets — if you can see it on both sides, circle it and cancel it.
Is a hidden factor lurking inside a polynomial? x^2 - a^2 always hides (x - a)(x + a). x^3 - a^3 hides (x - a)(x^2 + ax + a^2). x^2 - (a+b)x + ab hides (x - a)(x - b). If any of those hidden factors matches the denominator, the expression is about to collapse.
Is the numerator a multiple of the denominator? Sometimes the whole bottom divides the whole top cleanly. Divide and move on.

If any of the three checks fires, cancel first. Only if all three fail do you expand.

A visual way to see the move

The common factor $a$ appears once on top and once on bottom. Strike both, and what remains is $b + c$. No multiplication, no distribution, no term-by-term division — one cancellation, one line of work.

Try it yourself — spot and cancel

Drag $x$ along the number line. The full expression $x(x+3)/x$ and the cancelled form $x+3$ report the *same* number for every $x \neq 0$. The cancellation is not an approximation — it is the same function with one factor of $x$ removed from both sides.

The one pitfall — domain restrictions

When you cancel a from \tfrac{a(b+c)}{a}, you are dividing by a. Division by zero is undefined, so the simplification is only valid when a \neq 0. The cancelled form b + c is defined at a = 0; the original \tfrac{a(b+c)}{a} is not. They are equal as functions only on the domain a \neq 0.

In practice this matters in two places.

Limits. If you are taking a limit as a \to 0, you cancel first, then evaluate. The cancellation is valid for every a in the approach, so the limit of the cancelled form equals the limit of the original.
Domain-statement questions. If a JEE question asks for the domain of the original expression, do not collapse the a \neq 0 restriction into the simplified form. The domain is set by the original.

For ordinary algebraic simplification in the middle of a larger problem, you note the restriction and move on.

Common mistakes

"The a on the bottom cancels with both as on the top." No — you can only cancel one factor of a from each side at a time. \tfrac{a \cdot a \cdot b}{a} = ab, not b. Count factors.
"I can cancel across addition." \tfrac{a + b}{a} is not 1 + b. The a on the bottom only divides if it is a factor of the top, not a term. \tfrac{a + b}{a} = 1 + \tfrac{b}{a}, which is different. Spot the minus sign between "factor" and "term" — they are very different relationships.
"I'll cancel later, after I expand." Sometimes cancelling is easier to see before expansion — the hidden factor in x^2 - \pi^2 = (x-\pi)(x+\pi) is right there if you look, but it vanishes once you multiply out into x^2 - \pi^2. Expand only after cancellation, not before.

Why setters write it this way

JEE problems are time-constrained. A setter who wants a "60-second" question writes it so the 60-second route is obvious to a trained eye. The untrained student takes four minutes doing the hard version; the trained student takes 20 seconds doing the easy version. The problem is not harder for one or easier for the other — it is the same problem, seen through different lenses. Training the recognition is what the setter rewards.

When you see a common factor sitting on both sides of a fraction bar, you are looking at a gift. Take it.