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

Given $x^2 + \frac{1}{x^2} = 25$ and $x > 0$.

Using the identity $\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$:

$\left(x + \frac{1}{x}\right)^2 = 25 + 2 = 27$

Since $x > 0$:

$x + \frac{1}{x} = 3\sqrt{3}$

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Using the identity $\left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right)$:

$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$

$= (3\sqrt{3})^3 - 3(3\sqrt{3})$

$= 81\sqrt{3} - 9\sqrt{3}$

$= 72\sqrt{3}$

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Using the identity $\left(x^a + \frac{1}{x^a}\right)\left(x^b + \frac{1}{x^b}\right) = x^{a+b} + \frac{1}{x^{a+b}} + x^{a-b} + \frac{1}{x^{a-b}}$:

$x^5 + \frac{1}{x^5} = \left(x^3 + \frac{1}{x^3}\right)\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right)$

$= 72\sqrt{3} \times 25 - 3\sqrt{3}$

$= 1800\sqrt{3} - 3\sqrt{3}$

$= 1797\sqrt{3}$

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$x^7 + \frac{1}{x^7} = \left(x^5 + \frac{1}{x^5}\right)\left(x^2 + \frac{1}{x^2}\right) - \left(x^3 + \frac{1}{x^3}\right)$

$= 1797\sqrt{3} \times 25 - 72\sqrt{3}$

$= 44925\sqrt{3} - 72\sqrt{3}$

$= 44853\sqrt{3}$