The rational roots of the cubic equation ${{x}^{3}}+14k{{x}^{2}}+56kx-64{{k}^{3}}=0$ are in the
ratio 1:2:4. Find the possible value of k.
A. 0 only
B. 1 only
C. 2, 0
D. $-2,\ -1$

Hint: It is given that roots are in the ratio 1:2:4. So we will consider roots as P, 2P, 4P (which is in the 1:2:4 ratio)
Now we use sum of roots of equation in cubic as \[\dfrac{\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}}\]. to get equation---I
We also use product of roots in a cubic equation as \[\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{\text{3}}}}\] to get equation---II
Use both equations to calculate k and roots.

Complete step by step solution: Since the roots of the cubic equation ${{x}^{3}}+14k{{x}^{2}}+56kx-64{{k}^{3}}=0$ are in the
ratio 1:2:4. We consider the roots as P, 2P, 4P.
Therefore let roots of the cubic equation are P, 2P, 4P.
We know Sum of roots in cubic equation as \[=\dfrac{\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}}\].
$\therefore P+2P+4P=\dfrac{-14k}{1}$
$\therefore 7P=-14k$
$\therefore P=-2k$---------I
We also know product of roots in a cubic equation \[=\dfrac{-\text{constant term}}{\text{coefficient of }{{x}^{\text{3}}}}\]
$\left( P \right)\left( 2P \right)\left( 4P \right)=\dfrac{-\left( -64{{k}^{3}} \right)}{1}$
$\therefore 8{{P}^{3}}=64{{k}^{3}}$
$\therefore {{P}^{3}}=8{{k}^{3}}$--------II
But from eq I and II
$P=-2k$ (from eq I)
Substituting $P=-2k$ in eq --II we get
${{\left( -2k \right)}^{3}}=8{{k}^{3}}$
$-8{{k}^{3}}=8{{k}^{3}}$
$16{{k}^{3}}=0$
$\Rightarrow {{k}^{3}}=0$
$\Rightarrow k=0$
Since; $P=-2k$
$=-2\times 0$
$P=0$
$\therefore P=0$, $\therefore 2P=0$, $\therefore 4P=0$
All roots are 0 which is not possible
Since all the roots of this cubic equation are in the ratio 1:2:4, we cannot have one root as 0. As this will make other roots as also 0.
No option satisfy the answer (bonus question)

Note: One needs to remember that although we got one root as 0, it will make other roots as also 0. So ,all the roots are 0; But roots are in ratio 1:2:4. So we cannot have 0 as the root. So, none of the options satisfy the answer.