In order to simplify the radical, we will need to pull out common factors of possible perfect squares.

\(\displaystyle 12\sqrt{56} = 12\cdot \sqrt{8}\cdot \sqrt{7}=12\cdot \sqrt{4}\cdot \sqrt{2}\cdot \sqrt{7}\)

The expression becomes:  

\(\displaystyle 12\cdot 2\cdot \sqrt{2}\cdot \sqrt{7} = 24\sqrt{14}\)

The radical 14 does not have any common factors of perfect squares.

The answer is:  \(\displaystyle 24\sqrt{14}\)

Simplify by factoring both radicals by perfect squares.

\(\displaystyle \sqrt{45} = \sqrt{9}\cdot \sqrt{5} = 3\sqrt5\)

\(\displaystyle \sqrt{80} = \sqrt{16}\cdot \sqrt5 = 4\sqrt5\)

Replace the terms.

\(\displaystyle \sqrt{45}-\sqrt{80} = 3\sqrt5- 4\sqrt5\)

Combine like-terms.

The answer is:  \(\displaystyle -\sqrt5\)

Multiply the integers outside of the radical.

\(\displaystyle 6\sqrt{3}\cdot 5\sqrt{5}\cdot2\sqrt{10} =60\sqrt{3}\cdot \sqrt{5}\cdot\sqrt{10}\)

Multiply all the values inside the radicals to combine as one radical.

\(\displaystyle 60\sqrt{3}\cdot \sqrt{5}\cdot\sqrt{10} = 60\sqrt{150}\)

Rewrite the radical using factors of perfect squares.

\(\displaystyle 60\sqrt{150} = 60\sqrt{25} \cdot \sqrt6 = 60(5)\sqrt6\)

The answer is:  \(\displaystyle 300\sqrt6\)

The values of the radicals can be combined my multiplication.

\(\displaystyle 10\sqrt{1000}\cdot \sqrt{\frac{1}{8}} = 10\sqrt{\frac{1000}{8}} = 10\sqrt{125}\)

The value of \(\displaystyle \sqrt{125}\) can be factored by using known perfect squares as factors.

\(\displaystyle \sqrt{125} =\sqrt{25}\cdot \sqrt{5} = 5\sqrt5\)

Replace the term.

\(\displaystyle 10\sqrt{125} = 10\cdot 5\sqrt5= 50\sqrt5\)

The answer is:  \(\displaystyle 50\sqrt5\)

Rewrite each radical as common factors using known perfect squares.

\(\displaystyle 4[\sqrt{4}\cdot \sqrt2]\cdot [\sqrt{9}\cdot \sqrt3]\cdot[\sqrt{25}\cdot \sqrt2]\)

Simplify the expression.

\(\displaystyle 4[2\cdot \sqrt2]\cdot [3\cdot \sqrt3]\cdot[5\cdot \sqrt2]\)

\(\displaystyle 8 \cdot 15\sqrt3 \cdot 2 = 240\sqrt3\)

The answer is:  \(\displaystyle 240\sqrt3\)

A radical expression which is the \(\displaystyle n\)th root of a constant is in simplest form if and only if, when the radicand is expressed as the product of prime factors, no factor appears \(\displaystyle n\) or more times. Since \(\displaystyle \sqrt[3]{52}\) is a cube, or third, root, find the prime factorization of 52, and determine whether any prime factor appears three or more times.

52 can be broken down as

\(\displaystyle 52 = 2 \times 26\)

and further as

\(\displaystyle 52 = 2 \times 2 \times 13\)

No prime factor appears three or more times, so \(\displaystyle \sqrt[3]{52}\) is in simplest form.

First, apply the Quotient of Radicals Property to split the radical into a numerator and a denominator:

\(\displaystyle \sqrt[3]{\frac{2}{7}}\)

\(\displaystyle = \frac{\sqrt[3]{2}}{\sqrt[3]{7}}\)

Since we are dealing with cube, or third, roots, rationalize the denominator by multiplying both halves of the fraction by the least cube-root radical expression that would eliminate the radical in the denominator. To do this, note that 7 is a prime number. Therefore, to get a perfect cube, it is necessary to multiply both halves by \(\displaystyle \sqrt[3]{7 \cdot 7}\), and apply the Product of Radicals Property. The reason for this is made more apparent below:

\(\displaystyle \frac{\sqrt[3]{2} \cdot \sqrt[3]{7 \cdot 7}}{\sqrt[3]{7} \cdot \sqrt[3]{7 \cdot 7}}\)

\(\displaystyle =\frac{\sqrt[3]{2 \cdot 7 \cdot 7}}{\sqrt[3]{7 \cdot 7 \cdot 7}}\)

\(\displaystyle =\frac{\sqrt[3]{98}}{\sqrt[3]{7^{3}}}\)

\(\displaystyle =\frac{\sqrt[3]{98}}{7}\)

First, apply the Quotient of Radicals Property to split the radical into a numerator and a denominator:

\(\displaystyle \sqrt[3]{\frac{250}{8}}\)

\(\displaystyle =\frac{\sqrt[3]{250}}{\sqrt[3]{8}}\)

8 is a perfect cube - \(\displaystyle 8= 2^{3}\) - so the denominator can be simplified:

\(\displaystyle \frac{\sqrt[3]{250}}{\sqrt[3]{2^{3}}}\)

\(\displaystyle =\frac{\sqrt[3]{250}}{2}\)

To simplify the numerator, find the prime factorization of its radicand, 250, and look for any prime factors that appear three times:

\(\displaystyle 250 = 2 \times 125\)

\(\displaystyle = 2 \times 5 \times 25\)

\(\displaystyle = 2 \times 5 \times 5 \times 5\)

\(\displaystyle = 5 ^{3} \times 2\)

5 appears three times, so the numerator can be simplified by way of the Product of Radicals Property:

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

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

\(\displaystyle =\frac{ 5 \sqrt[3]{ 2 }}{2}\)

To rationalize a denominator, multiply all terms by the conjugate. In this case, the denominator is \(\displaystyle 6+\sqrt{27}\), so its conjugate will be \(\displaystyle 6-\sqrt{27}\).

So we multiply: \(\displaystyle \frac{1}{6+\sqrt{27}} \times \frac{6-\sqrt{27}}{6-\sqrt{27}} = \frac{6-\sqrt{27}}{36-27}\).

After simplifying, we get \(\displaystyle \frac{6-\sqrt{27}}{36-27} = \frac{3\cdot2-3\sqrt{3}}{9}=\frac{2-\sqrt{3}}{3}\).

The first step to simplifying a problem like this one is to convert all radicals to fractional exponents. Remember the following relationship:

\(\displaystyle \sqrt[n]{x}=x^{\frac{1}{n}}\)

Also keep in mind your exponent rules, especially this one:

\(\displaystyle (x^{m})^{n}=x^{mn}\)

Now, let's get started on this problem. First, we change that radical expression into something with fractional exponents instead.

\(\displaystyle f(x)=\frac{(\sqrt[4]{x^{3}})^{\frac{2}{5}}}{x^{6}}\)

\(\displaystyle f(x)=\frac{(x^{\frac{3}{4}})^{\frac{2}{5}}}{x^{6}}\)

Now we use our exponent rules to simplify the numerator.

\(\displaystyle f(x)=\frac{x^{(\frac{3}{4}\cdot\frac{2}{5})}}{x^{6}}=\frac{x^\frac{3}{10}}{x^{6}}\)

Finally, we simplify the entire fraction:

\(\displaystyle f(x)=\frac{x^{\frac{3}{10}}}{x^{\frac{60}{10}}}=x^{\frac{3}{10}}-x^{\frac{60}{10}}=x^{-\frac{53}{10}}\)

We can leave it like this, but it would be better to write it this way, without negative exponents:

\(\displaystyle f(x)=\frac{1}{x^{\frac{53}{10}}}\)