Question

Find the cube root of 217, correct to two decimal places.

Answer 1

Hint: For solving this type of questions we will first simplify the expression we have and later use the basic concept from the binomial theorem that the formulae for negative or fractional index are: $ {{\left( \text{1 }+\text{ x} \right)}^{n}}~=~1+n.x+\dfrac{n\left( n-1 \right)}{2!}x+\text{ }\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}x+......................\infty $ where $ \left| x \right|<1 $ .And use the basic exponential formula such as $ {{a}^{\dfrac{1}{n}}}=\sqrt[n]{a} $ and perform the calculations approximately very carefully.

Complete step by step answer:
Here according to the question we need to find the value of cube root of 217.
This can be mathematically expressed as $ \sqrt[3]{217} $ .
Since we know that $ {{a}^{\dfrac{1}{n}}}=\sqrt[n]{a} $ we can write this simply as
 $ \sqrt[3]{217}={{\left( 217 \right)}^{\dfrac{1}{3}}} $ .
Since we know that
  $ 216={{6}^{3}} $ .
$\Rightarrow$ We can write this as
  $ {{\left( 217 \right)}^{\dfrac{1}{3}}}={{\left( {{6}^{3}}+1 \right)}^{\dfrac{1}{3}}} $ .
Let us take out 6 for the sake of further simplifying the expression.
 $ {{\left( {{6}^{3}}+1 \right)}^{\dfrac{1}{3}}}=6{{\left( 1+\dfrac{1}{{{6}^{3}}} \right)}^{\dfrac{1}{3}}} $ .

Since we know that from the binomial theorem that
 $ {{\left( \text{1 }+\text{ x} \right)}^{n}}~=~1+n.x+\dfrac{n\left( n-1 \right)}{2!}x+\text{ }\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}x+......................\infty $ where $ \left| x \right|<1 $ .
$\Rightarrow$ We can simply use this formula and obtain the required value by placing $ x=\dfrac{1}{{{6}^{3}}} $ and $ n=\dfrac{1}{3} $ .
We will get $ {{\left( 1+\dfrac{1}{{{6}^{3}}} \right)}^{\dfrac{1}{3}}}=1+\dfrac{1}{3}\left( \dfrac{1}{{{6}^{3}}} \right)+\dfrac{\left( \dfrac{1}{3} \right)\left( \dfrac{1}{3}-1 \right)}{2!}\left( \dfrac{1}{{{6}^{3}}} \right)+.............................\infty $ .
We can simply write this as $ 6{{\left( 1+\dfrac{1}{{{6}^{3}}} \right)}^{\dfrac{1}{3}}}=6\left( 1+\dfrac{1}{3}\left( \dfrac{1}{{{6}^{3}}} \right)+\dfrac{\left( \dfrac{1}{3} \right)\left( \dfrac{1}{3}-1 \right)}{2!}\left( \dfrac{1}{{{6}^{3}}} \right)+.............................\infty \right) $ .
By performing further calculations we can say that
 $ \left( 1+\dfrac{1}{3}\left( \dfrac{1}{{{6}^{3}}} \right)+\dfrac{\left( \dfrac{1}{3} \right)\left( \dfrac{1}{3}-1 \right)}{2!}\left( \dfrac{1}{{{6}^{3}}} \right)+.............................\infty \right)=1.001 $ (Approximately)
 $ 6{{\left( 1+\dfrac{1}{{{6}^{3}}} \right)}^{\dfrac{1}{3}}}=6\left( 1+\dfrac{1}{3}\left( \dfrac{1}{{{6}^{3}}} \right)+\dfrac{\left( \dfrac{1}{3} \right)\left( \dfrac{1}{3}-1 \right)}{2!}\left( \dfrac{1}{{{6}^{3}}} \right)+.............................\infty \right)=6\left( 1.001 \right) $ .
$\Rightarrow$ Since, in the question, we have been asked to find the value of the cube root of 217 up to correct 2 decimals.
$\Rightarrow$ Here we will be having $ 6{{\left( 1+\dfrac{1}{{{6}^{3}}} \right)}^{\dfrac{1}{3}}}=6.009\approx 6.01 $ .
$\Rightarrow$ Hence we can conclude that $ \sqrt[3]{217} $ = 6.01.
$\Rightarrow$ Hence here the answer to the question is the value of the cube root of 217 is 6.01.

Note:
While solving questions of these type, one should keep in mind that the expansion relation,
 $ {{\left( x\text{ }+\text{ }y \right)}^{n}}~={{~}^{n}}{{\Sigma }_{r=0}}{{~}^{n}}{{C}_{r}}~{{x}^{n\text{ }\text{ }r~}}\cdot \text{ }{{y}^{r}}~{{=}^{n}}{{C}_{0}}~{{x}^{n\text{ }~}}\cdot \text{ }{{y}^{0~}}{{+}^{n}}{{C}_{1}}{{x}^{n-1}}\cdot {{y}^{1}}+....+{{\text{ }}^{n}}{{C}_{r}}~{{x}^{n\text{ }\text{ }r~}}\cdot \text{ }{{y}^{r~}}+....\text{ +}{{\text{ }}^{n}}{{C}_{n-1}}~x\text{ }\cdot \text{ }{{y}^{n\text{ }\text{ }1~}}+{{\text{ }}^{n}}{{C}_{n~}}\cdot \text{ }{{y}^{n}} $
is also valid for the negative/fractional index, this approximation can be simply written as $ {{\left( \text{1 }+\text{ x} \right)}^{n}}~=~1+n.x+\dfrac{n\left( n-1 \right)}{2!}x+\text{ }\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}x+......................\infty $ where $ \left| x \right|<1 $ . So, it would be better if one remembers the cubes of first 10 natural numbers so it will help to solve efficiently.