1Cademy - Simplifying $$\sqrt[3]{54} - \sqrt[3]{16}$$ and $$\sqrt[4]{48} + \sqrt[4]{243}$$

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Adding and Subtracting Like Square Roots

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Simplifying $\sqrt[3]{54} - \sqrt[3]{16}$ and $\sqrt[4]{48} + \sqrt[4]{243}$

Simplify two expressions where the higher roots must first be simplified to reveal like radicals that can be combined.

ⓐ $\sqrt[3]{54} - \sqrt[3]{16} = \sqrt[3]{2}$ :

Neither radicand is in simplified form. Rewrite each using its largest perfect cube factor: $54 = 27 \cdot 2$ and $16 = 8 \cdot 2$ . Apply the Product Property of cube roots:

$\sqrt[3]{27} \cdot \sqrt[3]{2} - \sqrt[3]{8} \cdot \sqrt[3]{2}$

Since $\sqrt[3]{27} = 3$ and $\sqrt[3]{8} = 2$ :

$3\sqrt[3]{2} - 2\sqrt[3]{2}$

Both terms now contain $\sqrt[3]{2}$ , so they are like radicals. Combine the coefficients: $3 - 2 = 1$ .

$\sqrt[3]{54} - \sqrt[3]{16} = \sqrt[3]{2}$

ⓑ $\sqrt[4]{48} + \sqrt[4]{243} = 5\sqrt[4]{3}$ :

Rewrite each radicand using its largest perfect fourth power factor: $48 = 16 \cdot 3$ and $243 = 81 \cdot 3$ . Apply the Product Property of fourth roots:

$\sqrt[4]{16} \cdot \sqrt[4]{3} + \sqrt[4]{81} \cdot \sqrt[4]{3}$

Since $\sqrt[4]{16} = 2$ and $\sqrt[4]{81} = 3$ :

$2\sqrt[4]{3} + 3\sqrt[4]{3}$

Both terms contain $\sqrt[4]{3}$ , making them like radicals. Combine the coefficients: $2 + 3 = 5$ .

$\sqrt[4]{48} + \sqrt[4]{243} = 5\sqrt[4]{3}$

In both parts, the original radicands appear different, but simplifying each radical by extracting the largest perfect $n$ th power factor reveals a common radical that allows the terms to be combined. This extends the simplify-then-combine strategy from square roots to higher-order roots.

Updated 2026-05-14

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