Question

If $x + \dfrac{1}{x} = 1$, then the value of $x^6 + \dfrac{1}{x^6}$ is

• A. –2
• B. –1
• C. 1
• D. 2

Hint:

For powers of $(x + 1/x)$, use recurrence relations: $x^n + 1/x^n = (x^{n-1} + 1/x^{n-1})(x + 1/x) - (x^{n-2} + 1/x^{n-2})$.

Answer 1

The Correct Option is D

Step 1: Given relation.
Let $x + \dfrac{1}{x} = 1$.
Step 2: Find $x^2 + \dfrac{1}{x^2}$.
Square both sides: \[ (x + \frac{1}{x})^2 = 1^2 \Rightarrow x^2 + \frac{1}{x^2} + 2 = 1 \] \[ \Rightarrow x^2 + \frac{1}{x^2} = -1 \] Step 3: Find $x^3 + \dfrac{1}{x^3}$.
\[ (x + \frac{1}{x})(x^2 + \frac{1}{x^2}) = x^3 + \frac{1}{x^3} + (x + \frac{1}{x}) \] \[ 1 \times (-1) = x^3 + \frac{1}{x^3} + 1 \Rightarrow x^3 + \frac{1}{x^3} = -2 \] Step 4: Find $x^6 + \dfrac{1}{x^6}$.
\[ (x^3 + \frac{1}{x^3})^2 = x^6 + \frac{1}{x^6} + 2 \] \[ (-2)^2 = x^6 + \frac{1}{x^6} + 2 \Rightarrow 4 = x^6 + \frac{1}{x^6} + 2 \] \[ \Rightarrow x^6 + \frac{1}{x^6} = 2 \] Wait! Let’s recheck sign consistency from step 2. Since $x^2 + \frac{1}{x^2} = -1$, recalculate step 3: \[ x^3 + \frac{1}{x^3} = (x + \frac{1}{x})(x^2 + \frac{1}{x^2}) - (x + \frac{1}{x}) = 1(-1) - 1 = -2 \] Now use step 4 again: \[ x^6 + \frac{1}{x^6} = (x^3 + \frac{1}{x^3})^2 - 2 = (-2)^2 - 2 = 4 - 2 = 2 \] Final verification: Correct value is 2.
Hence, answer = (D) 2.
Correction applied.