CAT 2025 Slot3 Quant: Algebra Identity

Step 1: Find $x + \frac{1}{x}$
We know $\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$.
Substituting the given value: $\left(x + \frac{1}{x}\right)^2 = 25 + 2 = 27$.
Since $x > 0$, $x + \frac{1}{x} = \sqrt{27} = 3\sqrt{3}$.

Step 2: Find $x^3 + \frac{1}{x^3}$
$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$
$= (3\sqrt{3})^3 - 3(3\sqrt{3}) = 81\sqrt{3} - 9\sqrt{3} = 72\sqrt{3}$.

Step 3: Find $x^4 + \frac{1}{x^4}$
$x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2$
$= (25)^2 - 2 = 625 - 2 = 623$.

Step 4: Final Calculation
$x^7 + \frac{1}{x^7} = (x^4 + \frac{1}{x^4})(x^3 + \frac{1}{x^3}) - (x + \frac{1}{x})$
$= (623)(72\sqrt{3}) - 3\sqrt{3}$
$= 44856\sqrt{3} - 3\sqrt{3} = \mathbf{44853\sqrt{3}}$.