Question

How do you simplify $ \dfrac{{x - 2}}{{{x^3} - 8}} $ ?

Answer 1

Hint: When the question asks to simplify an expression, we need to first factorize the polynomials in the numerator and denominator as much as possible. Factorization refers to writing a mathematical term as a product of several smaller terms of the same kind. These smaller terms are known as factors. So, try to factorize the denominator here by using some known identity.

Complete step by step answer:
(i)
We are given,
 $ \dfrac{{x - 2}}{{{x^3} - 8}} $
As we know that in rational expressions, denominator must not be equal to zero, so we can find restrictions from the original equation initially:
 $
  {x^3} - 8 \ne 0 \\
  {x^3} \ne 8 \\
  x \ne 2 \\
  $
So, $ x \ne 2 $ is the condition we have to remember till the end of the solution.
(ii)
Now, as we have the denominator is $ {x^3} - 8 $ which can also be written as:
 $ {x^3} - 8 = {x^3} - {2^3} $ [Since, we know that $ 2 $ is the cube-root of $ 8 $ as $ {2^3} = 8 $ ]
And we are asked to simplify the term. To simplify it, we need to factorize denominators. As we can clearly see that the denominator is in the form of $ {a^3} - {b^3} $ , we can apply the following identity to factorize the denominator.
 $ {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}) $
So, when we apply this identity to $ {x^3} - {2^3} $ , it becomes:
 $ {x^3} - {2^3} = (x - 2)({x^2} + 2x + {(2)^2}) $
Writing $ 4 $ in place of $ {2^2} $ , it will be:
 $ {x^3} - {2^3} = (x - 2)({x^2} + 2x + 4) $
(iii)
Substituting the factorized denominator in the question, we get:
 $ \dfrac{{x - 2}}{{(x - 2)({x^2} + 2x + 4)}} $
(iv)
Since, $ (x - 2) $ is present in both numerator and denominator, they can get cut with each other and we will obtain:
 $ \dfrac{1}{{({x^2} + 2x + 4)}} $
Hence, simplification of $ \dfrac{{x - 2}}{{{x^3} - 8}} $ will give us $ \dfrac{1}{{({x^2} + 2x + 4)}} $ with the condition $ x \ne 2 $ .

Note:
Do not forget to find the restriction for the value of the variable $ x $ as the denominator of a rational expression can never be zero. So always, the first step for the questions like this should be to find what value of $ x $, the denominator is going to be zero and then, remove that value from the range of $ x $ and write it with the final answer.