Question

Simplify the given expression : $\sqrt[4]{{81}} - 8\sqrt[3]{{216}} + 15\sqrt[3]{{32}} + \sqrt {225} $

Answer 1

Hint: We can simplify each of the terms separately. To simplify the terms inside the radicle, we can factorise the terms as prime numbers. Then we can simplify the radicals. After simplification, we can put it back into the expression. On further simplification, we will get the required answer.

Complete step-by-step answer:
We have the expression $\sqrt[4]{{81}} - 8\sqrt[3]{{216}} + 15\sqrt[3]{{32}} + \sqrt {225} $
Consider the 1st term $\sqrt[4]{{81}}$
We can factorise 81 as
 \[81 = 3 \times 3 \times 3 \times 3\]
We can write it as powers
 $ \Rightarrow 81 = {3^4}$
Now we can calculate the 4th root,
 $ \Rightarrow \sqrt[4]{{81}} = \sqrt[4]{{{3^4}}}$
Using \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}}\] , we get,
  $ \Rightarrow \sqrt[4]{{81}} = {3^{4 \times \dfrac{1}{4}}}$
On simplification we get,
 $ \Rightarrow \sqrt[4]{{81}} = 3$ … (1)
Now we can take the 2nd term $8\sqrt[3]{{216}}$
We can factorise 216 as
 $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$
We can write it as powers
 $ \Rightarrow 216 = {2^3} \times {3^3}$
Using \[{a^n} \times {b^n} = {(ab)^n}\] , we get,
 $ \Rightarrow 216 = {6^3}$
Thus, the term $8\sqrt[3]{{216}}$ will become.
 $ \Rightarrow 8\sqrt[3]{{216}} = 8 \times \sqrt[3]{{{6^3}}}$
Using \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}}\] , we get,
 $ \Rightarrow 8\sqrt[3]{{216}} = 8 \times {6^{3 \times \dfrac{1}{3}}}$
On simplification we get,
 $ \Rightarrow 8\sqrt[3]{{216}} = 8 \times 6$
 $ \Rightarrow 8\sqrt[3]{{216}} = 48$ …. (2)
Now we can consider the 3rd term, $15\sqrt[3]{{32}}$
We can write 32 as,
 $32 = 2 \times 2 \times 2 \times 2 \times 2$
We can write it as powers
 $ \Rightarrow 32 = {2^5}$
Now the term $15\sqrt[3]{{32}}$ will become,
 $ \Rightarrow 15\sqrt[3]{{32}} = 15 \times \sqrt[3]{{{2^5}}}$
Using \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}}\] , we get,
 $ \Rightarrow 15\sqrt[3]{{32}} = 15 \times {2^{5 \times \dfrac{1}{3}}}$
On simplification we get,
 $ \Rightarrow 15\sqrt[3]{{32}} = 15 \times 2 \times {2^{\dfrac{2}{3}}}$
Now we can write it as,
 $ \Rightarrow 15\sqrt[3]{{32}} = 15 \times 2 \times \sqrt[3]{{{2^2}}}$
On simplifying the powers, we get,
 $ \Rightarrow 15\sqrt[3]{{32}} = 30\sqrt[3]{4}$ …. (3)
Now consider the 4th term $\sqrt {225} $
We can write 225 as
 $225 = 5 \times 5 \times 3 \times 3$
We can write it as powers
 $ \Rightarrow 225 = {5^2} \times {3^2}$
Using \[{a^n} \times {b^n} = {(ab)^n}\] , we get,
 $ \Rightarrow 225 = {15^2}$
Thus $\sqrt {225} $ will become,
 $ \Rightarrow \sqrt {225} = \sqrt[2]{{{{15}^2}}}$
Using \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}}\] , we get,
 $ \Rightarrow \sqrt {225} = {15^{2 \times \dfrac{1}{2}}}$
On cancelling the powers, we get,
 $ \Rightarrow \sqrt {225} = 15$ …. (4)
On substituting (1), (2), (3) and (4), in the given expression, we get,
 $ \Rightarrow \sqrt[4]{{81}} - 8\sqrt[3]{{216}} + 15\sqrt[3]{{32}} + \sqrt {225} = 3 - 48 + 30\sqrt[3]{4} + 15$
 $ = - 30 + 30\sqrt[3]{4}$
Therefore, the simplified expression is $ - 30 + 30\sqrt[3]{4}$.

Note: We used the concept of square roots and cube roots to simplify the equation. The simplest method to find the nth root is by factorising the term and writing it in the exponential form. Then we can cancel the possible powers to get the required root. We use the concept that $\sqrt[n]{a} = {a^{\dfrac{1}{n}}}$ for cancelling the powers. We must understand that the number inside the square root is always positive and $\sqrt a $ is also positive. If ${x^2} = a$ , then x is given by, $x = \pm \sqrt a $ . Thus we obtain the positive and negative roots.