Question

If we have an expression as $x + \dfrac{1}{x} = 4$ then ${x^5} + \dfrac{1}{{{x^5}}} = ?$
A). $723$
B). $724$
C). $725$
D). $726$

Answer 1

Hint: In order to solve the equation for ${x^5} + \dfrac{1}{{{x^5}}}$, initiate with squaring the equation $x + \dfrac{1}{x} = 4$, solving it and multiplying the value with the obtained equation again. Similarly, multiply $\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)$ with $\left( {{x^3} + \dfrac{1}{{{x^3}}}} \right)$ in order to get the powers of x as 5. Solve the equations and get the result.

Complete step-by-step solution:
We are given with an equation $x + \dfrac{1}{x} = 4$. …….(1)
Starting with squaring both the sides of the above equation as ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we get:
${\left( {x + \dfrac{1}{x}} \right)^2} = {4^2}$
$ \Rightarrow {x^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2x\dfrac{1}{x} = 16$
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 16$
Subtracting both the sides by 2, we get:
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 - 2 = 16 - 2$
$ \Rightarrow \left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = 14$ ……..(2)
Multiplying the equation 1 with equation 2, we get:
$ \Rightarrow \left( {x + \dfrac{1}{x}} \right)\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = 4 \times 14$
Expanding the brackets by solving the operands, we get:
$ \Rightarrow x.{x^2} + x.\dfrac{1}{{{x^2}}} + \dfrac{1}{x}.{x^2} + \dfrac{1}{x}.\dfrac{1}{{{x^2}}} = 4 \times 14$
$ \Rightarrow {x^3} + \dfrac{1}{x} + x + \dfrac{1}{{{x^3}}} = 56$
Taking the cube of x common:
$ \Rightarrow \left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) + \left( {\dfrac{1}{x} + x} \right) = 56$
Substituting the value of $\left( {\dfrac{1}{x} + x} \right)$ in the above equation using the equation 1, we get:
$ \Rightarrow \left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) + 4 = 56$
Subtracting both sides by 4:
$ \Rightarrow \left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) + 4 - 4 = 56 - 4$
$ \Rightarrow \left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) = 52$ ……(3)
Since, we can know that we can get a power of x as 5, when we multiply ${x^3}$ and ${x^2}$, so we are multiplying equation 3 and equation 2, and we get:
$ \Rightarrow \left( {{x^3} + \dfrac{1}{{{x^3}}}} \right)\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = 52 \times 14$
Expanding the brackets, by multiplying the operands, we get:
$ \Rightarrow {x^3}.{x^2} + {x^3}.\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}.{x^2} + \dfrac{1}{{{x^3}}}.\dfrac{1}{{{x^2}}} = 52 \times 14$
Solving each operand’s:
$ \Rightarrow {x^5} + x + \dfrac{1}{x} + \dfrac{1}{{{x^5}}} = 728$
Taking the power of x of 5 common and other left operands as common, and we get:
$ \Rightarrow \left( {{x^5} + \dfrac{1}{{{x^5}}}} \right) + \left( {x + \dfrac{1}{x}} \right) = 728$
Substituting the value of $\left( {\dfrac{1}{x} + x} \right) = 4$ in the above equation from the equation 1, we get:
$ \Rightarrow \left( {{x^5} + \dfrac{1}{{{x^5}}}} \right) + 4 = 728$
Subtracting both sides by 4:
$ \Rightarrow \left( {{x^5} + \dfrac{1}{{{x^5}}}} \right) + 4 - 4 = 728 - 4$
$ \Rightarrow \left( {{x^5} + \dfrac{1}{{{x^5}}}} \right) = 724$
Since, we needed to find the value of ${x^5} + \dfrac{1}{{{x^5}}}$, so we got ${x^5} + \dfrac{1}{{{x^5}}} = 724$.
Hence, Option B is correct.

Note: We have written $x.{x^2} = {x^3}$ and ${x^3}.{x^2} = {x^5}$ as according to the law of radicals we know that powers are added in a product if the operands are common bases. It’s important to cross check the steps solved in order to avoid mistakes.