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

The correct option is $44853\sqrt{3}$.

To find the value of $x^7+\frac{1}{x^7}$, we can break the derivation down step-by-step using algebraic identities.

Step 1: Find the value of $x+\frac{1}{x}$

We are given:

$x^2+\frac{1}{x^2}=25$

Using the algebraic identity ${(a+b)}^2=a^2+b^2+2ab$:

${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2$

Substitute the given value:

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

Taking the square root on both sides (since $x>0$):

$x+\frac{1}{x}=\sqrt{27}=3\sqrt{3}$

Step 2: Find the value of $x^3+\frac{1}{x^3}$

Using the identity $a^3+b^3={(a+b)}^3-3ab(a+b)$:

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

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

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

$x^3+\frac{1}{x^3}=81\sqrt{3}-9\sqrt{3}=72\sqrt{3}$

Step 3: Find the value of $x^4+\frac{1}{x^4}$

Squaring $x^2+\frac{1}{x^2}=25$:

${(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2$

${25}^2=x^4+\frac{1}{x^4}+2$

$625=x^4+\frac{1}{x^4}+2\Rightarrow x^4+\frac{1}{x^4}=623$

Step 4: Combine the results to find $x^7+\frac{1}{x^7}$

Multiply $(x^3+\frac{1}{x^3})$ by $(x^4+\frac{1}{x^4})$:

$(x^3+\frac{1}{x^3})(x^4+\frac{1}{x^4})=x^7+\frac{1}{x}+x+\frac{1}{x^7}$

Rearranging the equation gives:

$x^7+\frac{1}{x^7}=(x^3+\frac{1}{x^3})(x^4+\frac{1}{x^4})-(x+\frac{1}{x})$

Substitute the calculated values into the formula:

$x^7+\frac{1}{x^7}=\left(72\sqrt{3}\right)\left(623\right)-3\sqrt{3}$

$x^7+\frac{1}{x^7}=44856\sqrt{3}-3\sqrt{3}=44853\sqrt{3}$