Question

If we have an expression as $ x - \dfrac{1}{x} = 5 $ , find the value of $ {x^3} - \dfrac{1}{{{x^3}}} $
A.167
B.169
C.140
D.160

Answer 1

Hint: In the given equation, the degree of x is one, but in the equation that we have to find out, the degree of x is three, so we have to calculate the cube of the given equation to convert it into an equation of degree three. Many identities can be used to solve the square of two or three numbers of terms. Using the appropriate identity, we can find the answer to the given question. After cubing, we simplify the equation obtained and put the value of the known terms if obtained.

Complete step-by-step answer:
We are given that $ x - \dfrac{1}{x} = 5 $
Cubing both sides, we get –
 $ {(x - \dfrac{1}{x})^3} = {(5)^3} $
We know $ {(a - b)^3} = {a^3} - {b^3} - 3ab(a - b) $
Using this identity in the above equation, we get –
 $
\Rightarrow {(x)^3} - {(\dfrac{1}{x})^3} - 3 \times x \times \dfrac{1}{x}(x - \dfrac{1}{x}) = 125 \\
\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} - 3(x - \dfrac{1}{x}) = 125 \\
  $
Put the value of $ x - \dfrac{1}{x} $ in the above equation,
 $
\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} - 3(5) = 125 \\
\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} = 125 + 15 \\
   \Rightarrow {x^3} - \dfrac{1}{{{x^3}}} = 140 \;
  $
So, the correct answer is “140”.

Note: We get the cube of a number by multiplying that number with itself three times, but there are some identities to find the cube of the sum of two terms. We know that the cube of the sum of two terms is equal to the sum of the cube of the first term, the cube of the third term, and the product of thrice the product of the first and second term and the sum of the two terms. That is –
 $ {(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) $
We can also write –
 $
  {[a + ( - b)]^3} = {a^3} + {( - b)^3} + 3a( - b)[a + ( - b)] \\
   \Rightarrow {(a - b)^3} = {a^3} - {b^3} - 3ab(a - b) \;
  $
We can also obtain this identity by multiplying the term $ (a - b) $ by itself thrice.