Question

How do you factor the sum or difference of two cubes ${{x}^{3}}-27$?

Answer 1

Hint: For this problem we need to calculate the factors of the given equation which is a difference of two cubes. In algebra we have the formula for the difference of two cubes as ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. So, we need to first convert each term in the given equation as cube value. For this we will use the factors of the number in the given equation and write it as the required form. Now we will compare the obtained value with ${{a}^{3}}-{{b}^{3}}$ and write the values of $a$, $b$. Now we will substitute those values in the algebraic formula and simplify it to get the required result.

Complete step by step solution:
Given that, ${{x}^{3}}-27$.
In the above equation we can observe that the first term is ${{x}^{3}}$ which is in already in cubic form. The second term we have in the given equation is $27$. When we prime factored the value $27$, then we will get $27=3\times 3\times 3={{3}^{3}}$. From this value, the given equation is modified as
$\Rightarrow {{x}^{3}}-27={{x}^{3}}-{{3}^{3}}$
Comparing the above equation with ${{a}^{3}}-{{b}^{3}}$, then we will get
$a=x$, $b=3$.
Using the algebraic formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ to simplify the given value, then we will get
$\begin{align}
  & \Rightarrow {{x}^{3}}-27=\left( x-3 \right)\left( {{x}^{2}}+\left( x \right)\left( 3 \right)+{{3}^{2}} \right) \\
 & \Rightarrow {{x}^{3}}-27=\left( x-3 \right)\left( {{x}^{2}}+3x+9 \right) \\
\end{align}$
Hence the factors of the given equation ${{x}^{3}}-27$ are $x-3$, ${{x}^{2}}+3x+9$.

Note: In the problem we have the difference of the two cubes which is mathematically represented by ${{a}^{3}}-{{b}^{3}}$. So, we have used the algebraic formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. If they have given the sum of the two cubes i.e., ${{a}^{3}}+{{b}^{3}}$, then we will use the algebraic formula which is ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$.