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# Simplify: $2\sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}-\sqrt[4]{16}$.

Last updated date: 20th Jun 2024
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Hint: We are given our equation in the form of radicals. We will first simplify each radical using $\sqrt[a]{b}={{\left( b \right)}^{\dfrac{1}{a}}}$, then we will factorize each radical, i.e (b) for each term we have. Then we will use ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$ to get simplified value. When we have all values, we will substitute them in the original equation and then using the rule of BODMAS we will simplify and get our required result.

Complete step by step answer:
We are asked to simplify:
$2\sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}-\sqrt[4]{16}$
To simplify this we have to simplify each radical. We know $\sqrt[a]{b}$ is defined as ${{a}^{th}}$ root of b and given as ${{\left( b \right)}^{\dfrac{1}{a}}}$ . Therefore, we have
$\Rightarrow \sqrt[a]{b}={{\left( b \right)}^{\dfrac{1}{a}}}$
So, we use this formula to simplify each radical.
$\sqrt[4]{81}={{\left( 81 \right)}^{\dfrac{1}{4}}}$
We know, $81=3\times 3\times 3\times 3={{3}^{4}}$ so we can substitute this as
\begin{align} & \sqrt[4]{81}={{\left( 81 \right)}^{\dfrac{1}{4}}} \\ & \Rightarrow {{\left( {{3}^{4}} \right)}^{\dfrac{1}{4}}} \\ & \text{As }{{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}} \\ & \Rightarrow {{3}^{4\times \dfrac{1}{4}}}=3 \\ & \Rightarrow \sqrt[4]{81}=3\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( 1 \right) \\ \end{align}
Now, let us consider next term,
\begin{align} & \sqrt[3]{216}={{\left( 216 \right)}^{\dfrac{1}{3}}} \\ & \text{As }216=6\times 6\times 6={{6}^{3}} \\ \end{align}
Now, we can substitute this and we will get $\Rightarrow \sqrt[3]{216}={{\left( {{6}^{3}} \right)}^{\dfrac{1}{3}}}$
Simplifying we will get,
$\Rightarrow \sqrt[3]{216}=6\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( 2 \right)$
Now, we will consider the term $\sqrt[5]{32}={{\left( 32 \right)}^{\dfrac{1}{5}}}$
On factorization we have 32 as $32=2\times 2\times 2\times 2\times 2={{2}^{5}}$
So substituting this here we will get
\begin{align} & \sqrt[5]{32}={{\left( 32 \right)}^{\dfrac{1}{5}}} \\ & \Rightarrow {{\left( {{2}^{5}} \right)}^{\dfrac{1}{5}}}=2 \\ \end{align}
After simplification we will get
$\Rightarrow \sqrt[5]{32}=2\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( 3 \right)$
Now, next we have
\begin{align} & \Rightarrow \sqrt{225}={{\left( 225 \right)}^{\dfrac{1}{2}}} \\ & \text{As }225=15\times 15={{15}^{2}} \\ \end{align}
So replacing 225 as ${{15}^{2}}$ we will get $\Rightarrow \sqrt{225}={{\left( {{15}^{2}} \right)}^{\dfrac{1}{2}}}=15$
$\Rightarrow \sqrt{225}=15\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( 4 \right)$
And lastly we will simplify $\sqrt[4]{16}$
\begin{align} & \Rightarrow \sqrt[4]{16}={{\left( 16 \right)}^{\dfrac{1}{4}}} \\ & \text{As }16=2\times 2\times 2\times 2={{2}^{4}} \\ \end{align}
So, we finally get it as
\begin{align} & \Rightarrow \sqrt[4]{16}={{\left( {{2}^{4}} \right)}^{\dfrac{1}{4}}}=2 \\ & \Rightarrow \sqrt[4]{16}=2\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( 5 \right) \\ \end{align}
Now, we will take the values from equations (1), (2), (3), (4) and (5) and substitute them in $2\sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}-\sqrt[4]{16}$ . Then, we will get
$\Rightarrow 2\times 3-8\times 6+15\times 2+15-2$
Simplifying using the BODMAS rule, we will get
\begin{align} & \Rightarrow 6-48+30+15-2 \\ & \Rightarrow 1 \\ \end{align}
So, we will get our answer as 1.
$2\sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}-\sqrt[4]{16}=1$

Note:
Remember ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$ so we can express ${{\left( {{3}^{4}} \right)}^{\dfrac{1}{4}}}={{3}^{4\times \dfrac{1}{4}}}={{3}^{\dfrac{4}{4}}}=3$ . Mistake like taking power ‘b’ over power ‘a’ may happen that is ${{\left( {{x}^{a}} \right)}^{b}}\ne {{x}^{ab}}$ so avoid such mistakes and try to solve properly as done above in the solution. We have to use the BODMAS rule, where first we have to multiply, then add and lastly subtract the terms while trying to simplify our last answer.