Find roots of polynomial $$ p(x) = 216d^3-1 $$
Answer
The roots of polynomial $ p(d) $ are:
$$ \begin{aligned}d_1 &= \frac{ 1 }{ 6 }\\[1 em]d_2 &= -\frac{ 1 }{ 12 }+\frac{\sqrt{ 3 }}{ 12 }i\\[1 em]d_3 &= -\frac{ 1 }{ 12 }- \frac{\sqrt{ 3 }}{ 12 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ d = \dfrac{ 1 }{ 6 } } $ is a root of polynomial $ 216d^3-1 $.
The Rational Root Theorem tells us that if the polynomial has a rational zero then it must be a fraction $ \dfrac{ \color{blue}{p}}{ \color{red}{q} } $, where $ p $ is a factor of the constant term and $ q $ is a factor of the leading coefficient.
The constant term is $ \color{blue}{ 1 } $, with factors of 1.
The leading coefficient is $ \color{red}{ 216 }$, with factors of 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 216 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1}\\[ 1 em] \pm \frac{ 1}{ 2}\\[ 1 em] \pm \frac{ 1}{ 3}\\[ 1 em] \pm \frac{ 1}{ 4}\\[ 1 em] \pm \frac{ 1}{ 6}\\[ 1 em] \pm \frac{ 1}{ 8}\\[ 1 em] \pm \frac{ 1}{ 9}\\[ 1 em] \pm \frac{ 1}{ 12}\\[ 1 em] \pm \frac{ 1}{ 18}\\[ 1 em] \pm \frac{ 1}{ 24}\\[ 1 em] \pm \frac{ 1}{ 27}\\[ 1 em] \pm \frac{ 1}{ 36}\\[ 1 em] \pm \frac{ 1}{ 54}\\[ 1 em] \pm \frac{ 1}{ 72}\\[ 1 em] \pm \frac{ 1}{ 108}\\[ 1 em] \pm \frac{ 1}{ 216}\\[ 1 em] \end{aligned} $$Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
We can see that $ p\left( \dfrac{ 1 }{ 6 } \right) = 0 $ so $ x = \dfrac{ 1 }{ 6 } $ is a root of a polynomial $ p(x) $.
To find remaining zeros we use Factor Theorem. This theorem states that if $ \dfrac{p}{q} $ is root of the polynomial then the polynomial can be divided by $ \color{blue}{qx − p} $. In this example we divide polynomial $ p $ by $ \color{blue}{ 6d-1 }$
$$ \frac{ 216d^3-1}{ 6d-1} = 36d^2+6d+1 $$Step 2:
The next rational root is $ d = \dfrac{ 1 }{ 6 } $
$$ \frac{ 216d^3-1}{ 6d-1} = 36d^2+6d+1 $$Step 3:
The solutions of $ 36d^2+6d+1 = 0 $ are: $ d = -\dfrac{ 1 }{ 12 }+\dfrac{\sqrt{ 3 }}{ 12 }i ~ \text{and} ~ d = -\dfrac{ 1 }{ 12 }-\dfrac{\sqrt{ 3 }}{ 12 }i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.