How do you factor \[125{{x}^{3}}-64{{y}^{3}}\]?

Hint: This question is from the topic of algebra. In this question, we have to find the factor of \[125{{x}^{3}}-64{{y}^{3}}\]. In solving this question, we will first find out the prime factorization of 125. After that, we will find out the prime factorization of 64. After that, by using the formula of \[{{a}^{3}}-{{b}^{3}}\] which is equal to \[\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we will solve the further equation. After solving the further question, we will get our answer.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to find out the factor of the term which is given in the question.
The term which we have to find the factor is:
\[125{{x}^{3}}-64{{y}^{3}}\]
Let us first find out the prime factorization of 125.
\[\begin{align}
  & 5\left| \!{\underline {\,
  125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
So, the prime factorization of 125 is \[5\times 5\times 5\times 1\]
Now, let us find out the prime factorization of 64.
\[\begin{align}
  & 2\left| \!{\underline {\,
  64 \,}} \right. \\
 & 2\left| \!{\underline {\,
  32 \,}} \right. \\
 & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
So, the prime factorization of 64 is \[2\times 2\times 2\times 2\times 2\times 2\times 1\].
Now, we can write the term \[125{{x}^{3}}-64{{y}^{3}}\] as
\[125{{x}^{3}}-64{{y}^{3}}=5\times 5\times 5\times 1\times {{x}^{3}}-2\times 2\times 2\times 2\times 2\times 2\times 1\times {{y}^{3}}\]
The above equation can also be written as
\[\Rightarrow 125{{x}^{3}}-64{{y}^{3}}=5\times 5\times 5\times {{x}^{3}}-2\times 2\times 2\times 2\times 2\times 2\times {{y}^{3}}\]
As we know that 2 multiplied by 2 is 4, or we can say \[2\times 2=4\], so we can write the above equation as
\[\Rightarrow 125{{x}^{3}}-64{{y}^{3}}=5\times 5\times 5\times {{x}^{3}}-4\times 4\times 4\times {{y}^{3}}\]
As we know that \[5\times 5\times 5\] can also be written as cube of 5 that is \[{{5}^{3}}\] and \[4\times 4\times 4\] can also be written as \[{{4}^{3}}\], so we can write the above equation as
\[\Rightarrow 125{{x}^{3}}-64{{y}^{3}}={{5}^{3}}\times {{x}^{3}}-{{4}^{3}}\times {{y}^{3}}\]
The above equation can also be written as
\[\Rightarrow 125{{x}^{3}}-64{{y}^{3}}={{\left( 5x \right)}^{3}}-{{\left( 4y \right)}^{3}}\]
Now, 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
\[\Rightarrow 125{{x}^{3}}-64{{y}^{3}}=\left( 5x-4y \right)\left( {{\left( 5x \right)}^{2}}+\left( 5x \right)\left( 4y \right)+{{\left( 4y \right)}^{2}} \right)\]
The above equation can also be written as
\[\Rightarrow 125{{x}^{3}}-64{{y}^{3}}=\left( 5x-4y \right)\left( 25{{x}^{2}}+20xy+16{{y}^{2}} \right)\]

Hence, we have found the factor of the term \[125{{x}^{3}}-64{{y}^{3}}\]. The factor of the term \[125{{x}^{3}}-64{{y}^{3}}\] is \[\left( 5x-4y \right)\left( 25{{x}^{2}}+20xy+16{{y}^{2}} \right)\].

Note: As we can see that this question is from the topic of algebra, so we should have a better knowledge in the topic of algebra. We should remember the following formula for solving this type of question:
\[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
For solving this type of question, we should know how to find prime factorization of any number.
We should know that \[5\times 5\times 5\] can also be written as \[{{5}^{3}}\] and \[4\times 4\times 4\] can also be written as \[{{4}^{3}}\].