Question

How many cube roots do $0.216$ have?

Answer 1

Hint: Here we have to find the cube roots of$0.216$. Firstly we will write the given number in rational form and then find the cube root of numerator and denominator separately. Then for the other cube root we will substitute the cube root of given value equal to $x$ and take the cube root we have already formed as the factor of that equation and solve for the other factor and get our desired answer.

Complete step-by-step answer:
We have to find the cube root of \[0.216\].
So changing the above value in rational form we will remove the decimal sign and divide the number by 10 power to the number of digit after the decimal and get,
\[0.216 = \dfrac{{216}}{{1000}}\]
Next, we will form the numerator and denominator in power of 3 as below,
\[ \Rightarrow 0.216 = \dfrac{{{6^3}}}{{{{10}^3}}}\]
Rewriting the equation, we get
\[ \Rightarrow 0.216 = {\left( {\dfrac{6}{{10}}} \right)^3}\]
Taking the cube root on both sides of the equation, we get
\[\begin{array}{l} \Rightarrow \sqrt[3]{{0.216}} = \sqrt[3]{{{{\left( {\dfrac{6}{{10}}} \right)}^3}}}\\ \Rightarrow \sqrt[3]{{0.216}} = {\left( {\dfrac{6}{{10}}} \right)^{\dfrac{1}{3} \times 3}}\end{array}\]
\[ \Rightarrow \sqrt[3]{{0.216}} = 0.6\]….\[\left( 1 \right)\]
Now, we will equate the cube root of \[0.216\] equal to \[x\].
\[x = \sqrt[3]{{0.216}}\]
Multiplying both side powers by 3 we get,
\[\begin{array}{l} \Rightarrow {x^3} = {0.216^{\dfrac{1}{3} \times 3}}\\ \Rightarrow {x^3} = 0.216\end{array}\]
Subtracting \[0.216\] from both the sides, we get
\[ \Rightarrow {x^3} - 0.216 = 0\]….\[\left( 2 \right)\]
As from above we can conclude that it will have 3 values which means we will get three cube root of \[0.216\] but we already know one cube root of the value from equation (1) which mens we have one factor of above equation.
 We have one factor as \[\left( {x - 0.6} \right)\].
Dividing equation (2) by \[\left( {x - 0.6} \right)\], we get,
\[\begin{array}{c}\dfrac{{{x^3} - 0.216}}{{x - 0.6}} = \dfrac{{\left( {x - 0.6} \right)\left( {{x^2} + 0.36 + 0.6x} \right)}}{{x - 0.6}}\\ = {x^2} + 0.36 + 0.6x\end{array}\]
So, we got our quotient as:
\[{x^2} + 0.6x + 0.36 = 0\]
Converting the decimal numbers into fractions, we get
\[ \Rightarrow {x^2} + \dfrac{6}{{10}}x + \dfrac{{36}}{{100}} = 0\]
Simplifying the fractions, we get
\[ \Rightarrow {x^2} + \dfrac{3}{5}x + \dfrac{9}{{25}} = 0\]
Taking L.C.M on RHS, we get
\[ \Rightarrow \dfrac{{25{x^2} + 15x + 9}}{{25}} = 0\]
Multiplying 25 on both the sides, we get
\[ \Rightarrow 25{x^2} + 15x + 9 = 0\]
Using quadratic formula, we get,
\[x = \dfrac{{ - \left( {15} \right) \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {25} \right) \times \left( 9 \right)} }}{{2 \times \left( {25} \right)}}\]
Simplifying the equation, we get
\[x = \dfrac{{ - 15 \pm \sqrt {225 - 900} }}{{50}}\]
Subtracting the terms, we get
\[\begin{array}{l}x = \dfrac{{ - 15 \pm \sqrt { - 675} }}{{50}}\\x = \dfrac{{ - 15 \pm 15\sqrt { - 3} }}{{50}}\end{array}\]
On solving the two terms and as we know \[\sqrt { - 1} = i\] we get,
\[x = \dfrac{{ - 15 + 15\sqrt 3 i}}{{50}} = \dfrac{{ - 3 + 3\sqrt 3 i}}{{10}} = - 0.3 + 0.3\sqrt 3 i\]….\[\left( 3 \right)\]
\[x = \dfrac{{ - 15 - 15\sqrt 3 i}}{{50}} = \dfrac{{ - 3 - \sqrt 3 i}}{{10}} = 0.3 - 0.3\sqrt 3 i\]…..\[\left( 4 \right)\]
From equation (1) (3) and (4) we get our three cube root as,
\[0.6, - 0.3 + 0.3\sqrt 3 i, - 0.3 - 0.3\sqrt 3 i\]

So, the three cube roots of \[0.216\] are \[0.6, - 0.3 + 0.3\sqrt 3 i, - 0.3 - 0.3\sqrt 3 i\].

Note:
All non-zero real numbers have exactly one real root and two complex conjugate roots but when it comes to non-zero complex numbers we get three complex cube roots. The cube root operation is not distributive with respect to addition and subtraction. One of the cube roots is termed as principal cube root. The concept of iota is used when getting the complex values.