Evaluate (216 × 6⁻⁵)³ × (36⁻²)⁻¹.

Step-by-step solution

1. 1

   Rewrite every number as a power of the same base. All the numbers involved are powers of $6$:

   $216=6^{3},\qquad 36=6^{2}$

   Once everything shares a base, only exponent arithmetic remains — no large numbers ever need to be computed.

2. 2

   Simplify inside the first bracket. Multiplying powers adds exponents:

   $216\cdot 6^{-5}=6^{3}\cdot 6^{-5}=6^{3-5}=6^{-2}$

3. 3

   Raise to the third power. A power of a power multiplies exponents:

   $\left(6^{-2}\right)^{3}=6^{-6}$

4. 4

   Simplify the second factor. Two nested exponents again multiply, and the two minus signs give a plus:

   $\left(36^{-2}\right)^{-1}=36^{2}=\left(6^{2}\right)^{2}=6^{4}$

5. 5

   Multiply the two results.

   $6^{-6}\cdot 6^{4}=6^{-2}=\frac{1}{6^{2}}=\frac{1}{36}$

   $\boxed{\dfrac{1}{36}}$

6. 6

   Check numerically. $216\cdot 6^{-5}=\tfrac{216}{7776}=\tfrac{1}{36}$; cubing gives $\tfrac{1}{46656}$; and $\left(36^{-2}\right)^{-1}=1296$. Then $\tfrac{1296}{46656}=\tfrac{1}{36}$ ✓ — the same answer with much larger intermediate numbers, which is exactly why the single-base route is preferable.