Rewrite the Base as a Power to Collapse a Fractional Exponent — the 8^(2/3) Move

On JEE Main you will see expressions like 8^{2/3} or 32^{-3/5} or 81^{3/4}, and the mark scheme expects you to write down the answer in one line. A student who reaches for a calculator or tries to compute the cube root of 8 from scratch has already lost time — there is a reflex move that collapses these expressions algebraically, in your head, and the reflex is worth training.

The move is: rewrite the base as a power of a smaller number, then apply the power-of-a-power law.

The drill

8^{2/3}. Recognise: 8 = 2^3. Substitute:

8^{2/3} = (2^3)^{2/3} = 2^{3 \cdot 2/3} = 2^2 = 4.

Three lines of algebra; no arithmetic beyond elementary-school multiplication. The fractional exponent 2/3 got absorbed into the inner exponent 3, leaving a clean whole number 2.

Compare with the naive approach: take 8^{2/3} to mean "cube root of 8, then square it," which is 2 then 4 — same answer, but you need to know that \sqrt[3]{8} = 2, which is the recognition step in disguise. The algebraic reframing makes the recognition step explicit and works for any base.

The general pattern

Whenever you see a^{p/q} where a is not prime, ask: can I rewrite a as b^q for some integer b? If yes, the fractional exponent will collapse:

a^{p/q} = (b^q)^{p/q} = b^{q \cdot p/q} = b^p.

The denominator q in the outer exponent cancels the inner exponent q inside the brackets. You are left with a whole-number exponent, which you can compute directly.

This is why problem-setters choose bases like 8, 16, 27, 32, 64, 81, 125, 243 and fractional exponents like 2/3, 3/5, 5/4. Each base is a cube, a fourth power, a fifth power — and the fractional exponent's denominator matches that power, so the collapse is clean.

The family you must memorise

Train your eye to see these bases as powers on instinct.

base	as a power	useful collapses
4	2^2	4^{1/2} = 2, 4^{3/2} = 8
8	2^3	8^{1/3} = 2, 8^{2/3} = 4, 8^{4/3} = 16
9	3^2	9^{1/2} = 3, 9^{3/2} = 27
16	2^4	16^{1/4} = 2, 16^{3/4} = 8, 16^{1/2} = 4
25	5^2	25^{1/2} = 5, 25^{3/2} = 125
27	3^3	27^{1/3} = 3, 27^{2/3} = 9
32	2^5	32^{1/5} = 2, 32^{2/5} = 4, 32^{3/5} = 8
64	2^6 = 4^3	64^{1/2} = 8, 64^{1/3} = 4, 64^{1/6} = 2
81	3^4	81^{1/4} = 3, 81^{3/4} = 27
125	5^3	125^{1/3} = 5, 125^{2/3} = 25
128	2^7	128^{1/7} = 2
243	3^5	243^{2/5} = 9, 243^{3/5} = 27
256	2^8 = 4^4	256^{1/4} = 4, 256^{1/8} = 2

Every one of these can appear in an exam question with a fractional exponent, and every one collapses in one line if you see the base-as-a-power decomposition. Carrying this table in your head is non-negotiable for JEE Mains.

Three JEE-style worked examples

Example 1. Simplify 32^{-3/5}.

Recognise: 32 = 2^5. Substitute:

32^{-3/5} = (2^5)^{-3/5} = 2^{5 \cdot (-3/5)} = 2^{-3} = \frac{1}{8}.

The 5 inside and the 5 in the denominator cancel. The negative sign flips the final answer to a reciprocal.

Example 2. Simplify \left(\dfrac{16}{81}\right)^{-3/4}.

Recognise: 16 = 2^4 and 81 = 3^4. Both are fourth powers, and the outer exponent has denominator 4 — perfect match.

\left(\frac{16}{81}\right)^{-3/4} = \frac{16^{-3/4}}{81^{-3/4}} = \frac{(2^4)^{-3/4}}{(3^4)^{-3/4}} = \frac{2^{-3}}{3^{-3}} = \frac{1/8}{1/27} = \frac{27}{8}.

The negative exponent flipped the whole thing upside down; the denominator matching both bases' powers made the collapse clean.

Example 3. Simplify 8^{2/3} + 27^{2/3} - 16^{3/4}.

Each term needs its own base decomposition:

8^{2/3} = (2^3)^{2/3} = 2^2 = 4.
27^{2/3} = (3^3)^{2/3} = 3^2 = 9.
16^{3/4} = (2^4)^{3/4} = 2^3 = 8.

Sum: 4 + 9 - 8 = 5.

An expression that looks complicated — three fractional exponents summed — collapses to simple integer arithmetic once each base is rewritten.

Why the collapse is so clean: the fractional exponent p/q is really asking "take the q-th root, then raise to the p-th." When the base is already a q-th power, the q-th root is trivial (it removes the exponent), and you are left to raise to the p-th. The algebraic rewrite makes this two-step operation explicit and avoids ever computing a root numerically.

When the rewrite doesn't help

Not every base is a clean power. 7^{2/3}, 15^{1/4}, 50^{3/5} — these cannot be rewritten, and fractional exponents on them genuinely need a calculator or leaving the answer in radical form. JEE problems rarely use such bases; when they do, the problem is usually setting up something other than a numerical evaluation (perhaps a limit or a comparison).

The recognition step is therefore a filter: first check whether the base is a perfect power of a small integer. If yes, collapse. If no, the problem wants something else from you — usually an algebraic identity or a comparison, not a numerical answer.

The decision tree for a^{p/q}

When you see a^{p/q}, run through this checklist:

Is a an integer power of a small integer (from the table above)? If yes, rewrite and collapse.
Is a itself a product that distributes, like (4 \cdot 9)^{1/2}? Distribute first, then collapse each factor.
Is a a variable or a non-perfect-power integer? Leave the answer as a^{p/q} or as a radical, depending on the context.

Ninety percent of JEE fractional-exponent questions hit case 1 or case 2. The reflex is to check case 1 first; it is the cleanest.

The reflex, in one line

When you see a fractional exponent on a non-prime base, rewrite the base as a power of something smaller and apply the power-of-a-power law. The denominator of the outer exponent will cancel the inner exponent, and the answer will fall out as an integer power of the smaller base.