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Converting surds from one order (index) to another involves changing the root of the expression while preserving its value. The most common conversion is changing between square roots (index 2) and higher-order roots (index greater than 2). Here's how you can convert surds from one order to another:

Converting from Higher-Order Root to Square Root:

To convert a higher-order root (√n) to a square root (√2), you can raise the expression to the power of the reciprocal of the index. In this case, the reciprocal of n is 1/n.

For example:

√3(8) can be converted to a square root by raising it to the power of 1/3:

√3(8) = (8)^(1/3) = ∛(8) = 2√2

Converting from Square Root to Higher-Order Root:

To convert a square root (√2) to a higher-order root (√n), you can raise the expression to the power of the desired index 'n'.

For example:

√2(18) can be converted to a cube root by raising it to the power of 3:

√2(18) = (18)^3 = 5832^(1/3) = ∛(5832) = 18∛2

It's important to note that not all surds can be converted to different orders while maintaining a simple expression. Some conversions may result in complex or irrational numbers.

Converting between Different Higher-Order Roots:

Converting between different higher-order roots involves finding a common index and then raising the expression accordingly. For example, to convert √5(32) to a cube root, you could first convert it to a fifth root and then to a cube root:

√5(32) = √5(32 * 1) = √5(32^(1/5))^5 = (32^(1/5))^5/5 = ∛(32) = 2∛2

When converting between different higher-order roots, you might need to use intermediate conversions to achieve the desired result.

**1. Convert \(\sqrt[3]{5}\) into a surd of order 6**

**Solution:**

\(\sqrt[3]{5}\)=\(5^{\frac{1}{3}}\)=\(5^{\frac{1 \times 2}{3 \times 2}}\)

\(=5^{\frac{2}{6}}=(5^{2})^{\frac{1}{6}}\)

\(=\sqrt[6]{25}\)

**2. Convert \(\sqrt{2}\) into a surd of order 12**

**Solution:**

\(\sqrt{2}=\sqrt[2 \times 6]{2^{6}}=\sqrt[12]{64}\)

**How to convert surds of different order into a surd of same order.**

**3. \(\sqrt[3]{4}, \sqrt[4]{3}\)
Find which is greater within the above numbers?**

**Solution:**

\(\sqrt[3]{4}, \sqrt[4]{3}\)

LCM of 3 and 4 is 12

\(\sqrt[3]{4}\)=\(\sqrt[3 \times 4]{4^{4}}\)

=\(\sqrt[12]{256}\)

\(\sqrt[4]{3}\)=\(\sqrt[4 \times 3]{3^{3}}\)

=\(\sqrt[12]{27}\)

So, \(\sqrt[3]{4}\) is greater