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:
$$44853\sqrt{3}$$

Step 1: Find the value of $x+\frac{1}{x}$
We are given:
$$x^2+\frac{1}{x^2}=25$$
Using the algebraic identity ${(a+b)}^2=a^2+b^2+2ab$, we can write:
$${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2$$
Substitute the given value $x^2+\frac{1}{x^2}=25$ into the equation:
$${(x+\frac{1}{x})}^2=25+2=27$$
Since $x>0$, the value of $x+\frac{1}{x}$ must be positive. Taking the square root on both sides:
$$x+\frac{1}{x}=\sqrt{27}=3\sqrt{3}$$

Step 2: Find the value of $x^3+\frac{1}{x^3}$
Using the identity $a^3+b^3={(a+b)}^3-3ab(a+b)$, we have:
$$x^3+\frac{1}{x^3}={(x+\frac{1}{x})}^3-3(x+\frac{1}{x})$$
Substituting $x+\frac{1}{x}=3\sqrt{3}$:
$$x^3+\frac{1}{x^3}={\left(3\sqrt{3}\right)}^3-3\left(3\sqrt{3}\right)$$
$$x^3+\frac{1}{x^3}=27\cdot 3\sqrt{3}-9\sqrt{3}=81\sqrt{3}-9\sqrt{3}=72\sqrt{3}$$

Step 3: Find the value of $x^4+\frac{1}{x^4}$
We can square $x^2+\frac{1}{x^2}$:
$${(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2$$
Substitute $x^2+\frac{1}{x^2}=25$:
$${25}^2=x^4+\frac{1}{x^4}+2$$
$$625=x^4+\frac{1}{x^4}+2$$
$$x^4+\frac{1}{x^4}=623$$

Step 4: Calculate the value of $x^7+\frac{1}{x^7}$
We multiply the expressions for the 3rd and 4th powers:
$$(x^3+\frac{1}{x^3})(x^4+\frac{1}{x^4})=x^7+\frac{1}{x^7}+(x+\frac{1}{x})$$
Rearranging the terms to isolate $x^7+\frac{1}{x^7}$:
$$x^7+\frac{1}{x^7}=(x^3+\frac{1}{x^3})(x^4+\frac{1}{x^4})-(x+\frac{1}{x})$$
Substituting the calculated values:
$$x^7+\frac{1}{x^7}=\left(72\sqrt{3}\right)\left(623\right)-3\sqrt{3}$$
Calculate $72\times 623$:
$$72\times 623=44856$$
So, the equation becomes:
$$x^7+\frac{1}{x^7}=44856\sqrt{3}-3\sqrt{3}=44853\sqrt{3}$$