$(2^{10} \cdot 3^{10}) / 6^{10}$ Simplifies to $1$ — Here's the Shortcut

You open a problem and stare at this:

\frac{2^{10} \cdot 3^{10}}{6^{10}}.

The first instinct is to panic. 2^{10} = 1024. 3^{10} = 59{,}049. 6^{10} = 60{,}466{,}176. Even on a calculator, you would have to multiply 1024 \cdot 59{,}049 (which is 60{,}466{,}176) and then divide by 60{,}466{,}176. The answer is 1, but the arithmetic is brutal.

There is a one-line shortcut. Spot the pattern — two powers with the same exponent — and regroup.

\frac{2^{10} \cdot 3^{10}}{6^{10}} = \frac{(2 \cdot 3)^{10}}{6^{10}} = \frac{6^{10}}{6^{10}} = 1.

Three steps, no arithmetic. Every step is a single application of an exponent law you already know.

The key recognition — a^n \cdot b^n = (ab)^n

The entire trick is built on one law, the power of a product rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}.

This is the fourth law in the list of the six exponent laws, and it is the most under-used one in exams. Every other law has obvious uses — the product law a^m \cdot a^n = a^{m+n}, the power-of-a-power (a^m)^n = a^{mn}, and so on. But a^n \cdot b^n = (ab)^n is the one that lets you combine different bases into a single base, which is exactly what you need when the exponents are already the same.

Why the law works: a^n \cdot b^n is n copies of a multiplied by n copies of b. Group them in pairs: (a \cdot b)(a \cdot b) \cdots (a \cdot b) with n pairs. That is (ab)^n. The commutativity of multiplication is what lets you interleave the as and bs.

When you spot that 2^{10} and 3^{10} both have the same exponent 10, you can fuse them into (2 \cdot 3)^{10} = 6^{10}. Now the numerator 6^{10} and the denominator 6^{10} match exactly, and they cancel to 1. Total arithmetic done: none.

The recognition pattern

Whenever you see a product or quotient of powers with matching exponents, the right move is to fuse the bases. Here are the three flavours:

Flavour 1 — fuse a product. a^n \cdot b^n = (ab)^n.

Flavour 2 — fuse a quotient. \dfrac{a^n}{b^n} = \left(\dfrac{a}{b}\right)^n.

Flavour 3 — fuse then cancel. If a \cdot b = c, then \dfrac{a^n \cdot b^n}{c^n} = \dfrac{(ab)^n}{c^n} = 1.

The original problem is Flavour 3 with a = 2, b = 3, c = 6.

The interactive recognition

Slide $n$ from $1$ to $12$. The numbers $2^n$, $3^n$, and $6^n$ all grow explosively — by $n = 12$, $6^{12}$ is more than two billion. But the ratio $(2^n \cdot 3^n)/6^n$ stays exactly $1$ for every $n$. The shortcut is not approximate; it is an identity.

A wider family of problems

Once you see the move, it works on any problem of this shape. A few examples you might actually meet in a JEE paper:

Example 1. Simplify \dfrac{4^{7} \cdot 25^{7}}{100^{7}}.

Because 4 \cdot 25 = 100, the numerator is (4 \cdot 25)^7 = 100^7, and the quotient is 1.

Example 2. Simplify \dfrac{15^{20}}{3^{20} \cdot 5^{20}}.

Because 3 \cdot 5 = 15, the denominator is (3 \cdot 5)^{20} = 15^{20}. The quotient is 1.

Example 3. Simplify 2^{10} \cdot 5^{10}.

Because 2 \cdot 5 = 10, the product is (2 \cdot 5)^{10} = 10^{10}. Simplified without computing 2^{10} or 5^{10} separately.

Example 4. Simplify \dfrac{8^{15}}{2^{15} \cdot 4^{15}}.

Because 2 \cdot 4 = 8, the denominator is 8^{15}. The quotient is 1.

Example 5. Evaluate \left(\dfrac{9}{16}\right)^{5} \cdot \left(\dfrac{16}{9}\right)^{5}.

Because the exponents match, fuse: \left(\dfrac{9}{16} \cdot \dfrac{16}{9}\right)^{5} = 1^{5} = 1.

In every one of these, brute-force expansion would be a disaster. The shortcut makes each of them a one-line answer.

Worked example — a slightly harder case

Problem. Simplify \dfrac{18^{10}}{2^{10} \cdot 3^{20}}.

Watch the exponents. 18^{10} has exponent 10. 2^{10} has exponent 10. 3^{20} has exponent 20 — different from the others. The fuse trick needs matching exponents, so you must first peel the 3^{20} apart.

3^{20} = (3^{2})^{10} = 9^{10}.

Why this works: power-of-a-power says (a^m)^n = a^{mn}. Reading it backwards, a^{mn} = (a^m)^n. So 3^{20} = 3^{2 \cdot 10} = (3^2)^{10} = 9^{10}. This is the law that lets you split or merge exponents as needed.

Now all three pieces have exponent 10, and we can fuse:

\frac{18^{10}}{2^{10} \cdot 9^{10}} = \frac{18^{10}}{(2 \cdot 9)^{10}} = \frac{18^{10}}{18^{10}} = 1.

Answer. 1.

The one-line takeaway

If you see a product or quotient of powers and you notice the exponents agree, regroup the bases using a^n \cdot b^n = (ab)^n or \dfrac{a^n}{b^n} = (a/b)^n. The arithmetic disappears and the simplification becomes obvious.

If the exponents do not agree, the trick is either to use the power-of-a-power law to make them match (as in the worked example), or to use the product law a^m \cdot a^n = a^{m+n} to combine powers of the same base. Between those three laws — product, quotient-of-same-base, and power-of-a-product — almost every exam exponent simplification is a short pattern-match away.

The reason this trick is worth practising until it becomes automatic is that exam time is short. A student who has to expand 2^{10} and 3^{10} by hand will lose five minutes. A student who recognises (2 \cdot 3)^{10} = 6^{10} answers in ten seconds. Over a full JEE paper, those savings add up to a letter-grade's worth of marks.