Question

If $\left(x^2 + \frac{1}{x^2}\right) = 25$ and $x>0$, then the value of $\left(x^7 + \frac{1}{x^7}\right)$ is

• A. \(44859\sqrt{3}\)
• B. \(44853\sqrt{3}\)
• C. \(44850\sqrt{3}\)
• D. \(44856\sqrt{3}\)

Hint:

For expressions like $x^n + \dfrac{1}{x^n}$, first find $x + \dfrac{1}{x}$ using the given data, then use the recurrence \[ S_n = S_1 S_{n-1} - S_{n-2}, \] instead of expanding powers directly. It saves a lot of time and reduces mistakes in exams.

Answer 1

The Correct Option is B

Approach: Never jump straight to the 7th power. Climb a ladder \(x+\tfrac1x \to x^2+\tfrac1x^2 \to x^3+\tfrac1x^3 \to \dots\) and combine two lower powers to reach the one you want. The key recurrence is \(\left(x^a+\tfrac1{x^a}\right)\left(x^b+\tfrac1{x^b}\right)=\left(x^{a+b}+\tfrac1{x^{a+b}}\right)+\left(x^{a-b}+\tfrac1{x^{a-b}}\right).\)

Step 1: Get \(x+\tfrac1x\). Since \(\left(x+\tfrac1x\right)^2=x^2+\tfrac1{x^2}+2=25+2=27\), and \(x>0\) forces a positive value, \(x+\tfrac1x=\sqrt{27}=3\sqrt3.\)

Step 2: \(x^3+\tfrac1{x^3}=\left(x+\tfrac1x\right)\left(x^2+\tfrac1{x^2}\right)-\left(x+\tfrac1x\right)=3\sqrt3(25)-3\sqrt3=72\sqrt3.\)

Step 3: \(x^6+\tfrac1{x^6}=\left(x^3+\tfrac1{x^3}\right)^2-2=(72\sqrt3)^2-2=72^2\cdot3-2=15552-2=15550.\)

Step 4: \(x^5+\tfrac1{x^5}=\left(x^3+\tfrac1{x^3}\right)\left(x^2+\tfrac1{x^2}\right)-\left(x+\tfrac1x\right)=72\sqrt3(25)-3\sqrt3=1800\sqrt3-3\sqrt3=1797\sqrt3.\)

Step 5: Finally use \(x^7+\tfrac1{x^7}=\left(x^6+\tfrac1{x^6}\right)\left(x+\tfrac1x\right)-\left(x^5+\tfrac1{x^5}\right).\)

\[x^7+\tfrac1{x^7}=15550\cdot3\sqrt3-1797\sqrt3=46650\sqrt3-1797\sqrt3=44853\sqrt3.\]

Final answer: \(\;x^7+\dfrac1{x^7}=44853\sqrt3.\)

Answer 2

Step 1: Find \(x + \frac{1}{x}\). We are given: \[ x^2 + \frac{1}{x^2} = 25. \] Recall: \[ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}. \] So, \[ \left(x + \frac{1}{x}\right)^2 = 25 + 2 = 27 \quad \Rightarrow \quad x + \frac{1}{x} = \sqrt{27} = 3\sqrt{3}, \] since \(x > 0\) implies \(x + \frac{1}{x} > 0\). Let \[ S_1 = x + \frac{1}{x} = 3\sqrt{3}. \]  
Step 2: Use identities / recurrence to build higher powers. We use the standard recurrence: \[ S_n = x^n + \frac{1}{x^n}, \quad \text{and} \quad S_n = S_1 S_{n-1} - S_{n-2}, \text{ for } n \ge 2. \] We already know: \[ S_1 = 3\sqrt{3}, \quad S_2 = x^2 + \frac{1}{x^2} = 25. \] Now compute step by step. 
\(\underline{For }\) \(n = 3\): \[ S_3 = S_1 S_2 - S_1 = (3\sqrt{3})(25) - 3\sqrt{3} = 72\sqrt{3}. \] \(\underline{For } \)\(n = 4\): \[ S_4 = S_1 S_3 - S_2 = (3\sqrt{3})(72\sqrt{3}) - 25 = 3 \cdot 72 \cdot 3 - 25 = 648 - 25 = 623. \]\( \underline{For }\) \(n = 5\): \[ S_5 = S_1 S_4 - S_3 = (3\sqrt{3})(623) - 72\sqrt{3} = (1869 - 72)\sqrt{3} = 1797\sqrt{3}. \] \(\underline{For }\) \(n = 6\): \[ S_6 = S_1 S_5 - S_4 = (3\sqrt{3})(1797\sqrt{3}) - 623 = 3 \cdot 1797 \cdot 3 - 623 = 16173 - 623 = 15550. \] \(\underline{For }\) \(n = 7\): \[ S_7 = S_1 S_6 - S_5 = (3\sqrt{3})(15550) - 1797\sqrt{3} = (46650 - 1797)\sqrt{3} = 44853\sqrt{3}. \] Thus, \[ x^7 + \frac{1}{x^7} = 44853\sqrt{3}. \]