How do you factorize $ 54{{x}^{3}}-2{{y}^{3}} $ .

Hint: We are asked to factorize $ 54{{x}^{3}}-2{{y}^{3}} $ , we will use step by step simplification to do so. First, we will take the common multiple which could be common to both the terms, after that we see how many terms are there and which identity can be used. We will use $ {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) $ , we will also $ \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) $ to learn about how we can factorize different kind of terms.

Complete step by step answer:
We are given an expression, $ 54{{x}^{3}}-2{{y}^{3}} $ and we are asked to factorize the expression. Factorizing an expression means that we have to split the expression into a simpler form. For example, if we consider 2x+4, we can see that 2x and 4 have 2 in common, so we can take it out. So, we will get,
2x+4 = 2(x+2).
Now, we see that it cannot be simplified further, hence, this is the simplest form of this expression.
Now, let us consider another example, $ {{x}^{2}}-{{y}^{2}} $ , to simplify it, we use the identity $ \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) $ . Considering x as a and y as b, we get,
The factorization of $ {{x}^{2}}-{{y}^{2}} $ will be $ {{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right) $ .
Now, to factorize a term, we will follow certain steps.
Step 1: Factor out greatest common factor first if possible.
In our problem, we have the expression $ 54{{x}^{3}}-2{{y}^{3}} $ , we can see that 54 can be written as $ 2\times 27 $ and we can see that in the term $ 2{{y}^{3}} $ also we have a 2, hence we can take out 2 as greatest common factor. So, we will get,
 $ 54{{x}^{3}}-2{{y}^{3}}=2\left( 27{{x}^{3}}-{{y}^{3}} \right) $
Step 2: Count the number of terms, if we have 2 terms, we usually use $ \left( {{a}^{2}}-{{b}^{2}} \right) $ or $ \left( {{a}^{3}}-{{b}^{3}} \right) $ or $ \left( {{a}^{3}}+{{b}^{3}} \right) $ . $ \left( {{a}^{2}}-{{b}^{2}} \right) $ is defined as $ \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) $ , $ \left( {{a}^{3}}-{{b}^{3}} \right) $ is defined as $ {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) $ and $ \left( {{a}^{3}}+{{b}^{3}} \right) $ is defined as $ \left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) $ .
In our expression, we have 2 terms as $ 54{{x}^{3}}-2{{y}^{3}}=2\left( 27{{x}^{3}}-{{y}^{3}} \right).........(1) $ .
We have $ \left( 27{{x}^{3}}-{{y}^{3}} \right) $ , so we will use the identity $ \left( {{a}^{3}}-{{b}^{3}} \right) $ and apply this to find the value of a and b.
Now, as we have $ 27{{x}^{3}} $ , where we know that 27 can be written as $ 3\times 3\times 3 $ and $ {{x}^{3}}=x\times x\times x $ .
So, $ 27{{x}^{3}}={{\left( 3x \right)}^{3}} $ .
Similarly, $ {{y}^{3}}={{\left( y \right)}^{3}} $ .
So, we have a = 3x and b = y.
Hence, using $ {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) $ , we will get,
 $ \left( 27{{x}^{3}}-{{y}^{3}} \right)=\left( 3x-y \right)\left( {{\left( 3x \right)}^{2}}+{{y}^{2}}+3xy \right) $
On simplifying, we get,
 $ \left( 27{{x}^{3}}-{{y}^{3}} \right)=\left( 3x-y \right)\left( 9{{x}^{2}}+{{y}^{2}}+3xy \right) $
Now, substituting this in (1), we will get,
 $ 54{{x}^{3}}-2{{y}^{3}}=2\left[ \left( 3x-y \right)\left( 9{{x}^{2}}+{{y}^{2}}+3xy \right) \right] $
So, we get the factor of $ 54{{x}^{3}}-2{{y}^{3}} $ as $ 2\left[ \left( 3x-y \right)\left( 9{{x}^{2}}+{{y}^{2}}+3xy \right) \right] $ .

Note:
Remember that $ 2{{y}^{3}} $ is not the same as $ {{\left( 2y \right)}^{3}} $ . In $ 2{{y}^{3}} $ , the cube is only for y, but in $ {{\left( 2y \right)}^{3}} $ , the cube is for 2y. When we multiply variables with each other like, $ x\times {{x}^{2}} $ , then their powers get added up and we will get it as $ {{x}^{3}} $ . Also keep in mind that only like terms can be added together.