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
Show Hint
- \(44859\sqrt{3}\)
- \(44853\sqrt{3}\)
- \(44850\sqrt{3}\)
- \(44856\sqrt{3}\)
The Correct Option is B
Approach Solution - 1
To find the value of \(x^7 + \frac{1}{x^7}\), we start with the given equation:
\(x^2 + \frac{1}{x^2} = 25\)
Our goal is to express \(x^7 + \frac{1}{x^7}\) in terms of known values. We will use successive powers.
Let us first calculate \(x + \frac{1}{x}\) from \(x^2 + \frac{1}{x^2} = 25\).
We have the identity:
\(\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2\)
Substituting the given:
\(\left(x + \frac{1}{x}\right)^2 = 25 + 2 = 27\)
Thus, \(x + \frac{1}{x} = \sqrt{27} = 3\sqrt{3}\) (since \(x > 0\)).
Now, find \(x^3 + \frac{1}{x^3}\) using the identity:
\(x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right)\)
Substituting values:
\(x^3 + \frac{1}{x^3} = (3\sqrt{3})(25) - 3\sqrt{3} = 75\sqrt{3} - 3\sqrt{3} = 72\sqrt{3}\)
To find \(x^6 + \frac{1}{x^6}\), use:
\(\left(x^3 + \frac{1}{x^3}\right)^2 = x^6 + \frac{1}{x^6} + 2\)
Substituting:
\((72\sqrt{3})^2 = x^6 + \frac{1}{x^6} + 2\)
\(x^6 + \frac{1}{x^6} = (72^2 \times 3) - 2 = 15552 - 2 = 15550\)
Finally, find \(x^7 + \frac{1}{x^7}\) using:
\(x^7 + \frac{1}{x^7} = \left(x^6 + \frac{1}{x^6}\right)\left(x + \frac{1}{x}\right) - \left(x^5 + \frac{1}{x^5}\right)\)
We need \(x^5 + \frac{1}{x^5}\), which can be found using the previous identity:
\(x^5 + \frac{1}{x^5} = \left(x^3 + \frac{1}{x^3}\right)\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right)\)
Substituting values gives:
\(x^5 + \frac{1}{x^5} = (72\sqrt{3})(25) - (3\sqrt{3}) = 1800\sqrt{3} - 3\sqrt{3} = 1797\sqrt{3}\)
Substitute this in the equation for \(x^7 + \frac{1}{x^7}\):
\(x^7 + \frac{1}{x^7} = (15550)(3\sqrt{3}) - (1797\sqrt{3}) = 46650\sqrt{3} - 1797\sqrt{3} = 44853\sqrt{3}\)
Thus, the correct answer is \(44853\sqrt{3}\).
Approach Solution -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, \[ x^7 + \frac{1}{x^7} = 44853\sqrt{3}. \]
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