Question

If ${x^3} + \dfrac{1}{{{x^3}}} = 110$, then what is the value of $x + \dfrac{1}{x}$ ?
A. $5$
B. $10$
C. $15$
D. None of these.

Answer 1

Hint: In this question, we are given the value of ${x^3} + \dfrac{1}{{{x^3}}}$ and are asked to find the value of $x + \dfrac{1}{x}$. Start by using the formula of ${(a + b)^3}$ and put the values given in the question. Now, find the required value using the hit and trial method. Put the values given in the question one by one. If the equation is satisfied, the particular value is the answer. If not, the answer is none of these.

Formula used: ${(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$ or ${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$

Complete step-by-step answer:
We are given- ${x^3} + \dfrac{1}{{{x^3}}} = 110$
Using the formula ${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$, let $a = x$, and $b = \dfrac{1}{x}$.
Now we have, ${(x + \dfrac{1}{x})^3} = {x^3} + \dfrac{1}{{{x^3}}} + 3(x + \dfrac{1}{x})$
Substituting ${x^3} + \dfrac{1}{{{x^3}}} = 110$,
$ \Rightarrow {(x + \dfrac{1}{x})^3} = 110 + 3(x + \dfrac{1}{x})$
For simplicity, let us assume $(x + \dfrac{1}{x}) = y$. Now, we have
$ \Rightarrow {y^3} = 110 + 3y$
Shifting RHS to the other side,
$ \Rightarrow {y^3} - 3y - 110 = 0$ ……………... (1)
Now to find the value of $y$, we will use hit and trial method. We will try putting all the values given in the options. If any value satisfies with the above equation, it will be considered correct. And if not, then our answer will be (D) None of these.
Starting with the option (A) $5$ and putting in equation (1).
$ \Rightarrow {5^3} - 15 - 110 = 0$
$ \Rightarrow 125 - 15 - 110 = 0$
$ \Rightarrow 0 = 0$
Since $y = 5$ satisfies the equation (1) , it is our required answer.
But for practice, let’s try with option (2) as well.
Putting $y = 10$ in equation (1),
$ \Rightarrow {10^3} - 30 - 110 \ne 0$
Since $y = 10$ does not satisfy the equation (1), it is not the required answer.
Now, $x + \dfrac{1}{x} = y$ and $y = 5$.
Therefore, $x + \dfrac{1}{x} = 5$.

Option A is the correct answer.

Note: We have assumed $\left( {x + \dfrac{1}{x}} \right) = y$ only for the purpose of simplification. The student can solve the answer directly also. However, it will be a little confusing and tedious. It will go like-
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^3} - 3\left( {x + \dfrac{1}{x}} \right) = 110$
Now, since we have to find the value of $\left( {x + \dfrac{1}{x}} \right)$ , we will put $\left( {x + \dfrac{1}{x}} \right) = 5$ as we are starting from option (A).
$ \Rightarrow {5^3} - 15 = 110$
$ \Rightarrow 125 - 15 = 110$
$ \Rightarrow 110 = 110$
Here, you can observe that writing $\left( {x + \dfrac{1}{x}} \right)$ again and again is tedious. Due to this, we prefer assuming $\left( {x + \dfrac{1}{x}} \right) = y$.