Question

How do you factor \[\dfrac{1}{8}{{x}^{3}}-\dfrac{1}{27}{{y}^{3}}\]?

Answer 1

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

Complete step by step answer:
We are asked to factorize the expression \[\dfrac{1}{8}{{x}^{3}}-\dfrac{1}{27}{{y}^{3}}\].
We know that 8 is the cube of 2, and 27 is the cube of 3. Using this in the above expression, it can be written as \[{{\left( \dfrac{1}{2} \right)}^{3}}{{x}^{3}}-{{\left( \dfrac{1}{3} \right)}^{3}}{{y}^{3}}\].
Using the algebraic property \[{{a}^{n}}{{b}^{n}}={{\left( ab \right)}^{n}}\], we can simplify this expression as
\[\Rightarrow {{\left( \dfrac{1}{2}x \right)}^{3}}-{{\left( \dfrac{1}{3}y \right)}^{3}}\]
 This expression is of the form of difference 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 \[a=\dfrac{1}{2}x\And b=\dfrac{1}{3}y\]. substituting the values of a and b for this question in the expansion of difference of cubes, we get
\[\Rightarrow {{\left( \dfrac{1}{2}x \right)}^{3}}-{{\left( \dfrac{1}{3}y \right)}^{3}}=\left( \dfrac{1}{2}x-\dfrac{1}{3}y \right)\left( {{\left( \dfrac{1}{2}x \right)}^{2}}+\left( \dfrac{1}{2}x \right)\left( \dfrac{1}{3}y \right)+{{\left( \dfrac{1}{3}y \right)}^{2}} \right)\]
We know that the square of 3 is 9, and the square of 2 is 4. substituting this value above expression and, simplifying the above expression, we get
\[\Rightarrow {{\left( \dfrac{1}{2}x \right)}^{3}}-{{\left( \dfrac{1}{3}y \right)}^{3}}=\left( \dfrac{1}{2}x-\dfrac{1}{3}y \right)\left( \dfrac{1}{4}{{x}^{2}}+\dfrac{1}{6}xy+\dfrac{1}{9}{{y}^{2}} \right)\]

Thus, the factored form of the given expression is \[\left( \dfrac{1}{2}x-\dfrac{1}{3}y \right)\left( \dfrac{1}{4}{{x}^{2}}+\dfrac{1}{6}xy+\dfrac{1}{9}{{y}^{2}} \right)\].

Note: To solve these types of questions, we must know the algebraic expansions of different expressions like difference of square which is algebraically expressed as\[{{a}^{2}}-{{b}^{2}}\] and its expansion is \[(a+b)(a-b)\], addition of cubes which is algebraically expressed as \[{{a}^{3}}+{{b}^{3}}\] and its expansion is \[(a+b)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\].
We can factorize such expressions by finding the values of a and b, and then substitute it in the expansion of the expressions.