Simplifying a cube root with a perfect cube factor

Key takeaway: This question checks whether you can separate perfect cube factors from a radical and simplify variable exponents correctly.

Key idea

To simplify

$\sqrt[3]{81x^3y^2},$

look for factors that are perfect cubes. A cube root is simplified by extracting any factor whose exponent is a multiple of 3.

Here, $81 = 27\cdot 3$, and $27$ is a perfect cube. Also, $x^3$ is a perfect cube because $\sqrt[3]{x^3}=x$.

Step-by-step simplification

Rewrite the radicand as

$81x^3y^2 = 27\cdot 3\cdot x^3\cdot y^2.$

Now take the cube root of each perfect cube factor:

$\sqrt[3]{27x^3\cdot 3y^2} = \sqrt[3]{27}\,\sqrt[3]{x^3}\,\sqrt[3]{3y^2}.$

Since

$\sqrt[3]{27}=3 \quad \text{and} \quad \sqrt[3]{x^3}=x,$

the expression becomes

$3x\sqrt[3]{3y^2}.$

So the correct choice is (d).

Why the other options are tempting

The choices that place a 9 or 27 outside the radical usually come from treating $81$ as if it were a perfect cube itself. It is not. Only $27$ can come out cleanly as a cube root. Another common mistake is pulling $y^2$ outside, but $y^2$ is not a perfect cube, so it must stay inside the radical.

A clean way to check your answer is to cube the outside factor and confirm it matches the factors that were removed from the radical. The leftover part should contain no exponent that is a multiple of 3.

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Pitfalls the pros know 👇 A frequent error is to simplify $\sqrt[3]{81}$ as 9, but that would be the square root thinking pattern, not the cube root pattern. For cube roots, you must look for factors like 8, 27, 64, 125, and variable exponents divisible by 3. Another mistake is splitting the radical incorrectly and trying to take $\sqrt[3]{y^2}$ outside. Since 2 is not a multiple of 3, that term must stay inside. Check exponents carefully before you extract anything.

What if the problem changes? If the problem were

$\sqrt[3]{54x^6y^5},$

you would factor it as $27\cdot 2\cdot x^6\cdot y^3\cdot y^2$, giving

$3x^2y\sqrt[3]{2y^2}.$

If the variable exponent were $x^4$ instead of $x^3$, only one $x$ would come out because $x^4=x^3\cdot x$. The same cube-root method works, but the number of factors that leave the radical depends on how many full groups of 3 are present.

Tags: Cube root simplification, Perfect cubes, Radical factors