Question

What is the approximate value of the cube root of the number $9$?
A) $2.08$
B) $2.19$
C) $2.34$
D) $2.51$

Answer 1

Cube of a number can be found by using binomial formula. From this we can eliminate small quantities if, because we are asked for the approximate value, not the actual value. For using a binomial equation, we have to find the perfect cube nearer to $9$.

Formula used:
For any $a,b$ and natural number $n$, we have
${(a + b)^n} = {a^n} + 3{C_1}{a^{n - 1}}b + 3{C_2}{a^{n - 2}}{b^2} + ... + 3{C_{n - 1}}{a^{}}{b^{n - 1}} + {b^{^n}}$
Here $C$ denotes the combination and ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Particularly for $n = 3$,
${(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$

Complete step-by-step answer:
We are asked to find the cube root approximation of $9$.
For we can find the perfect cube nearer to $9$.
We can see ${2^3} = 8$.
At the same time ${3^3} = 27$.
So we understand the approximate cube root of $9$ is very close to $2$.
Therefore we can let $\sqrt[3]{9} = 2 + \in$, where $\in$ is a very small quantity.
We have the equation, ${(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$
Substituting for $a = 2,b = \in$ we get,
${(2 + \in )^3} = {2^3} + 3 \times {2^2} \times \in + 3 \times 2 \times { \in ^2} + { \in ^3}$
Since $\in$ is a very small quantity, ${ \in ^2},{ \in ^3} \to 0$. So we can neglect the last two terms in the above equation.
$\Rightarrow {(2 + \in )^3} = {2^3} + 3 \times {2^2} \times \in$
Simplifying we get,
$\Rightarrow {(2 + \in )^3} = 8 + 12 \in$
We had let $\sqrt[3]{9} = 2 + \in$.
So, ${(2 + \in )^3} = {(\sqrt[3]{9})^3} = 9$
$\Rightarrow 9 = 8 + 12 \in$
Subtracting $8$ from both sides we have,
$\Rightarrow 1 = 12 \in$
Dividing both sides by $12$ we get,
$\Rightarrow \in = \dfrac{1}{{12}} \sim 0.08$
So we get the value of $\in$ approximately equal to $0.08$.
This gives $2 + \in \sim 2 + 0.08 = 2.08$
$\sqrt[3]{9} = 2 + \in \Rightarrow \sqrt[3]{9} \sim 2.08$
So, the approximate value of cube root of $9$is $2.08$.
$\therefore$ The answer is option A.

Note: Here we used this method since $9$ is not a perfect cube. For perfect cubes we can find cube roots by prime factorisation and grouping the numbers. Also there are other methods as well for finding roots. If the number given was lesser than its nearest perfect cube we can use the equation of ${(a - b)^3}$.