Question

How do you factor the following differences of two cubes \[54{{x}^{3}}-16{{y}^{3}}\]?

Answer 1

Hint: This question is from the topic of algebra. In this question, we have to factor out the term given in the question. In solving this question, we will take the common factor 2. After that, we will convert the given term in simple cubic form. After solving the further process, we will get our answer.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to factor out the term which is given in the question. The term given in the question is:
\[54{{x}^{3}}-16{{y}^{3}}\]
We can see in the above term that there are 54 and 16. And, we know that 54 can also be written as 27 multiplied by 2 and 16 can also be written as 8 multiplied by 2. So, we can write the term \[54{{x}^{3}}-16{{y}^{3}}\] as
\[54{{x}^{3}}-16{{y}^{3}}=27\times 2\times {{x}^{3}}-8\times 2\times {{y}^{3}}\]
Now, we can see in the right side of the above equation that there is a common factor that is 2. Then, we can write the above equation as
\[\Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left( 27\times {{x}^{3}}-8\times {{y}^{3}} \right)\]
As we know that the cube of 3 is 27 and the cube of 2 is 8. So, we can write the above equation as
\[\Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left( {{3}^{3}}\times {{x}^{3}}-{{2}^{3}}\times {{y}^{3}} \right)\]
The above equation can also be written as
\[\Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left( {{\left( 3\times x \right)}^{3}}-{{\left( 2\times y \right)}^{3}} \right)\]
The above equation can also be written as
\[\Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left( {{\left( 3x \right)}^{3}}-{{\left( 2y \right)}^{3}} \right)\]
Using the formula: \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we can write the above equation as
\[\Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left( \left( 3x \right)-\left( 2y \right) \right)\left( {{\left( 3x \right)}^{2}}+\left( 3x \right)\left( 2y \right)+{{\left( 2y \right)}^{2}} \right)\]
The above equation can also be written as
\[\Rightarrow 54{{x}^{3}}-16{{y}^{3}}=2\left( 3x-2y \right)\left( 9{{x}^{2}}+6xy+4{{y}^{2}} \right)\]
Now, we have factor out the term \[54{{x}^{3}}-16{{y}^{3}}\]. We got the answer as \[2\left( 3x-2y \right)\left( 9{{x}^{2}}+6xy+4{{y}^{2}} \right)\].

Note: We should have a better knowledge in the topic of algebra to solve this type of question easily. We should remember the formula of \[{{a}^{3}}-{{b}^{3}}\]. The formula for \[{{a}^{3}}-{{b}^{3}}\] is in the following:
\[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
We should know that the 27 is a cube of 3 and 8 is a cube of 2.