Estimation of cube root - class-VIII

First multiply and divide by $1,000,000,$ we get
$\sqrt[3]{\dfrac{9\times 1000,000}{1000,000}}$
$\sqrt[3]{9,000,000} \div 100$
Now find the closest cube root of $9,000,000.$
$208^3 = 8,998,912,$ therefore we can say that $\sqrt[3]{9}$ $\sim$ $\dfrac{208}{100} \sim 2.08$

$\sqrt[3] {\dfrac{12\times 1000}{1000}}$
$\sqrt[3]{12000} \div 10$
Now find the closest cube root of $12000.$
$22^3 = 10648,$ therefore we can say that $\sqrt[3]{12}$ $\sim$ $\dfrac{22}{10} = 2.2$

$\displaystyle \sqrt[3]{3\left ( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right )}=2$ 
Let us assume $\left( \sqrt [ 3 ]{ x } -\frac { 1 }{ \sqrt [ 3 ]{ x }  }  \right) $=a
So, $\sqrt [ 3 ]{ 3a } =2$
Taking  cube both sides,
$3a=8$
$a=\frac { 8 }{ 3 } $
So, $\left( \sqrt [ 3 ]{ x } -\frac { 1 }{ \sqrt [ 3 ]{ x }  }  \right) =\frac { 8 }{ 3 } $

$\sqrt[3]{768}=\sqrt[3]{2^3\times2^3\times2\times2\times3}=2\times2\sqrt[3]{12}=4\sqrt[3]{12}$

We need to find value of $\sqrt[3]{45}$
Take, $n = 45$, choose any starting value of $x$.
So, $3^3$ is $27 < 45$
So, $x = 3$
$x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
$x _\text{next} = $ $\dfrac{2}{3}3+\dfrac{45}{3\times 3^2}$
$x _\text{next} = 3.6666$
So, the nearest whole number for the cube root $45$ is $4$.

We need to find $\sqrt[3]{1333}$
Take, $n = 1333$, choose any starting value of $x$.
So, $11^3$ is $1331 < 1333$
So, $x = 11$
$x _\text{next}$ $=$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
$x _\text{next}$ $=$ $\dfrac{2}{3}11+\dfrac{1331}{3\times 11^2}$
$x _\text{next}$ $= 10.999 $    ....(1)
Assume $x = 10.99$
$x _\text{next} =$ $\dfrac{2}{3}10.99+\dfrac{1331}{3\times 10.99^2}$
$x _\text{next} = 10.99$     ....(2)
Since we are getting $10.99$ in (1) and (2)
So, the approximate value of $\sqrt[3]{1333}$ $= 10.99$

We use the Babylonian Algorithm for cube roots here

We need to find value of $\sqrt[3]{1290}$
Take, $n = 1290$, choose any starting value of $x$.
So, $10^3$ is $1000 < 1290$
So, $x = 10$
$x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
$x _\text{next} =$ $\dfrac{2}{3}10+\dfrac{1290}{3\times 10^2}$
$x _\text{next} =  10.966$
So, the nearest whole number for the cube root $1290$ is $11$.

We need to find value of $\sqrt[3]{6860}$
Take, $n = 6860$, choose any starting value of $x$.
So, $19^3$ is $6859 < 6860$
So, $x = 19$
$x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
$x _\text{next} = $ $\dfrac{2}{3}19+\dfrac{6860}{3\times 19^2}$
$x _\text{next} = 19.00085 $    ....(1)
So, the approximate value of $\sqrt[3]{6860}$ nearest hundredth place is $19.00$.

We need to find value of $\sqrt[3]{823}$
Take, $n = 823$, choose any starting value of $x$.
So, $9^3$ is $729 < 823$
So, $x = 9$
$x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
$x _\text{next} =$ $\dfrac{2}{3}9+\dfrac{823}{3\times 9^2}$
$x _\text{next} = 9.386$
So, the nearest whole number for the cube root $823$ is $9$.

We need to find value of $\sqrt[3]{2486}$
Take, $n = 2486$, choose any starting value of $x$.
So, $13^3$ is $2197 < 2486$
So, $x = 13$
$x _\text{next} =$ $\dfrac{2}{3}x+\dfrac{n}{3x^2}$
$x _\text{next} =$ $\dfrac{2}{3}13+\dfrac{2486}{3\times 13^2}$
$x _\text{next} = 13.56$
So, the nearest whole number for the cube root $2486$ is $14$.