Factorize the equation $50{{a}^{3}}-2a$.
(a) $2a\left( 5a+1 \right)\left( 5a+6 \right)$
(b) $2a\left( a+1 \right)\left( 1-a \right)$
(c) $2a\left( 5a+1 \right)\left( 5a-1 \right)$
(d) $2a\left( a+1 \right)\left( a+1 \right)$

Hint: We start solving the problem by taking the common factor from both the terms. We then use the one of the law of exponents ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$ to proceed through the problem. We then use the property that $\left( {{x}^{2}}-{{y}^{2}} \right)=\left( x-y \right)\left( x+y \right)$ to proceed further into the problem. We then make the necessary arrangements to get the required result of factorization of the given equation.

Complete step by step solution:
According to the problem, we need to factorize the equation $50{{a}^{3}}-2a$.
$\Rightarrow 50{{a}^{3}}-2a=\left( 2a.25{{a}^{2}} \right)-\left( 2a.1 \right)$.
We can see that $2a$ is a common factor in both terms. Let us take that factor as common from both terms.
So, we get $50{{a}^{3}}-2a=2a\left( 25{{a}^{2}}-1 \right)$.
$\Rightarrow 50{{a}^{3}}-2a=2a\left( {{5}^{2}}.{{a}^{2}}-1 \right)$ ---(1).
From the laws of exponents, we know that ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$. Let us use this result in equation (1).
$\Rightarrow 50{{a}^{3}}-2a=2a\left( {{\left( 5a \right)}^{2}}-{{\left( 1 \right)}^{2}} \right)$ ---(2).
We can see that ${{\left( 5a \right)}^{2}}-{{\left( 1 \right)}^{2}}$ resembles with ${{x}^{2}}-{{y}^{2}}$ which is equal to $\left( x+y \right)\left( x-y \right)$. Let us use this result in equation (2).
So, we get $50{{a}^{3}}-2a=2a\left( \left( 5a+1 \right)\left( 5a-1 \right) \right)$.
$\Rightarrow 50{{a}^{3}}-2a=2a\left( 5a+1 \right)\left( 5a-1 \right)$.
So, we have found the factorization of the equation $50{{a}^{3}}-2a$ as $2a\left( 5a+1 \right)\left( 5a-1 \right)$.
The correct option for the given problem is (c).

Note: Whenever we get this type of problem, we first try to take common factors present in the given terms of the independent variable. We can also solve this problem by finding the zeros of the given equation, as we can see that the form of the equation is similar to the polynomial. We should not neglect 2 in the factorization as we need to get a similar form on expanding the result of factorization. Similarly, we can expect problems to find the zeros and degree of the polynomial.