Question

How do you factor \[10{{x}^{3}}-270\]?

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 \[{{a}^{3}}-{{b}^{3}}=(a-b)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]. The constants in the given expression are 10 and 270, these are not perfect cubes. But, if we take 10 as a common factor from both of them, we get 1 and 27, these are cubes of 1 and 3.

Complete step by step answer:
We are asked to factorize the expression \[10{{x}^{3}}-270\].
As 10 and 270 have the highest common factor as 10 taking 10 common from both, we get \[10\left( {{x}^{3}}-27 \right)\].
We know that 1 is the cube of 1, and 27 is the cube of 3. Using this in the above expression, it can be written as \[10\left( {{x}^{3}}-{{3}^{3}} \right)\].
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 \[a=x\And b=3\]. substituting the values of a and b for this question in the expansion of difference of cubes, we get
\[10\left( {{x}^{3}}-{{3}^{3}} \right)=10(x-3)\left( {{x}^{2}}+x3+{{3}^{2}} \right)\]
We know that the square of 3 is 9, substituting this value above and, simplifying the above expression, we get
\[10\left( {{x}^{3}}-{{3}^{3}} \right)=10(x-3)\left( {{x}^{2}}+3x+9 \right)\]

Thus, the factored form of the given expression is \[10(x-3)\left( {{x}^{2}}+3x+9 \right)\].

Note: As we did for this example, sometimes we need to take a common factor from both of the constant values. As the given values may not be perfect cubes/ squared of a number. The expansion of the form \[{{a}^{2n}}+{{b}^{2n}}\], here n is a positive integer. These can not be further factorized.