Question

How do you factor \[125{{x}^{3}}+\dfrac{1}{125}\]?

Answer 1

Hint: To solve the given question, we need to know the algebraic expansion of \[{{a}^{3}}+{{b}^{3}}\]. The expressions of these forms are called the addition of cubes. This expression is expanded as \[{{a}^{3}}+{{b}^{3}}=(a+b)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\]. For the given expression we have \[a=5x\And b=\dfrac{1}{5}\]. We will solve the given problem by substituting the values of a and b in the expansion.

Complete step by step answer:
We are asked to factorize the expression \[125{{x}^{3}}+\dfrac{1}{125}\].
We know that 125 is the cube of 5. Using this in the above expression, it can be written as \[{{5}^{3}}{{x}^{3}}+\dfrac{1}{{{5}^{3}}}\]. Using the algebraic property \[{{a}^{n}}{{b}^{n}}={{\left( ab \right)}^{n}}\], we can simplify this expression as
\[\Rightarrow {{\left( 5x \right)}^{3}}+\dfrac{1}{{{5}^{3}}}\]
This expression is of the form of addition of cubes, we know that the expansion of the form \[{{a}^{3}}+{{b}^{3}}=(a+b)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\]. Here, we have \[a=5x\And b=\dfrac{1}{5}\]. substituting the values of a and b for this question in the expansion of addition of cubes, we get
\[{{\left( 5x \right)}^{3}}+\dfrac{1}{{{5}^{3}}}=\left( 5x+\dfrac{1}{5} \right)\left( {{\left( 5x \right)}^{2}}-(5x)\left( \dfrac{1}{5} \right)+{{\left( \dfrac{1}{5} \right)}^{2}} \right)\]
We know that the square of 5 is 25, substituting this value above and, simplifying the above expression, we get
\[{{\left( 5x \right)}^{3}}+\dfrac{1}{{{5}^{3}}}=\left( 5x+\dfrac{1}{5} \right)\left( 25{{x}^{2}}-x+\dfrac{1}{25} \right)\]

Thus, the factored form of the given expression is \[\left( 5x+\dfrac{1}{5} \right)\left( 25{{x}^{2}}-x+\dfrac{1}{25} \right)\].

Note: To solve these types of questions, we must know the algebraic expansions of different expressions like difference of square \[{{a}^{2}}-{{b}^{2}}\], difference of cubes which is algebraically expressed as \[{{a}^{3}}-{{b}^{3}}\].
The expansions for these expressions are as follows, the difference of square \[{{a}^{2}}-{{b}^{2}}\] is expanded as \[\left( a+b \right)\left( a-b \right)\]. The difference of cubes is expanded as \[{{a}^{3}}-{{b}^{3}}=(a-b)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]. To use these expansions, we first have to find the value of a and b, then substitute it in the expansion.