Question

The real number ${\left(\sqrt[3]{\sqrt{75}-\sqrt{12}}\right)}^{-2}$ when expressed in the simplest form is equal to ?

Answer 1

Hint: In this question, we need to write simplify the given real number ${\left(\sqrt[3]{\sqrt{75}-\sqrt{12}}\right)}^{-2}$ and express it in the simplest form possible. We will use a law of radical, to write the given term into the simplest form of exponent. The law states that, if $n$ is a positive integer and a is any real number, then $\sqrt[n]{a^x}={\left(a\right)}^{{\displaystyle \frac{x}{n}}}$ . Then, we will substitute the values and determine the simplest form of the given exponential function.

Complete step by step solution:
Here, we need to convert ${\left(\sqrt[3]{\sqrt{75}-\sqrt{12}}\right)}^{-2}$ in the simplest form.
Expressing in simplest radical form is nothing but simplifying the radical into the simplest form with no more square roots, cube roots, etc. left to find. In other words, a number under a radical is indivisible by a perfect square other than $1$.
Now, we first factorize the numbers $75$ and $12$ using the prime factorization method.
So, we get, $75=3\times 5\times 5$ and $12=3\times 2\times 2$ .
Hence, simplifying the expression ${\left(\sqrt[3]{\sqrt{75}-\sqrt{12}}\right)}^{-2}$, we get,
$\Rightarrow {\left(\sqrt[3]{5\sqrt{3}-2\sqrt{3}}\right)}^{-2}$
Now, adding up the like terms and simplifying the expression, we get,
$\Rightarrow {\left(\sqrt[3]{3\sqrt{3}}\right)}^{-2}$
Now, we have to evaluate the cube root. Now, we know that $3\sqrt{3}=\sqrt{3}\times \sqrt{3}\times \sqrt{3}$. So, we can express $3\sqrt{3}$ as ${\left(\sqrt{3}\right)}^3$. Hence, we get,
$\Rightarrow {\left(\sqrt{3}\right)}^{-2}$
Now, This is of the form ${\left(a^m\right)}^{{\displaystyle \frac{1}{n}}}$. But we can rewrite the expression using the laws of indices and powers $a^{\left({\displaystyle \frac{m}{n}}\right)}={\left(a^m\right)}^{{\displaystyle \frac{1}{n}}}$ . So, we get,
$\Rightarrow {\left(\sqrt{3}\right)}^{-2}={\left[{\left(\sqrt{3}\right)}^2\right]}^{-1}$
So, evaluating the power of the base, we get,
$\Rightarrow {\left(3\right)}^{-1}$
Now, evaluating the negative exponent and simplifying the expression, we get,
$\Rightarrow \left({\displaystyle \frac{1}{3}}\right)$
Hence, the required answer of the expression ${\left(\sqrt[3]{\sqrt{75}-\sqrt{12}}\right)}^{-2}$ is $\left({\displaystyle \frac{1}{3}}\right)$ .

Note:
 In this question, it is important to note here that we used a law of the radical to solve this form i.e., a radical represents a fractional exponent in which the numerator of the fractional exponent is the power of the base and the denominator of the fractional exponent is the index of the radical.