$\text{If}x\text{is a positive real number such that}{\mathrm{ x}}^8+{\left(\frac{1}{x}\right)}^8=47,\text{then the value of}{\mathrm{ x}}^9+{\left(\frac{1}{x}\right)}^9\text{is}$

The correct answer is $34\sqrt{5}$.

We are given that $x$ is a positive real number ($x>0$) and:
$x^8+\frac{1}{x^8}=47$
We want to find the value of $x^9+\frac{1}{x^9}$.

Step 1: Simplify the exponents to find $x+\frac{1}{x}$
Using the algebraic identity ${(a^2+b^2)}^2=a^4+b^4+2a^2{b}^2$, we can write:
${(x^4+\frac{1}{x^4})}^2=x^8+\frac{1}{x^8}+2$
Substituting the given value of $x^8+\frac{1}{x^8}=47$:
${(x^4+\frac{1}{x^4})}^2=47+2=49$
Since $x>0$, the term $x^4+\frac{1}{x^4}$ must be positive. Taking the square root:
$x^4+\frac{1}{x^4}=7$

Similarly, we can find $x^2+\frac{1}{x^2}$:
${(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2=7+2=9$
Taking the positive square root:
$x^2+\frac{1}{x^2}=3$

Now, we find $x+\frac{1}{x}$:
${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2=3+2=5$
Since $x>0$:
$x+\frac{1}{x}=\sqrt{5}$

Step 2: Determine the value of $x^3+\frac{1}{x^3}$
Using the cubic identity $a^3+b^3={(a+b)}^3-3ab(a+b)$:
$x^3+\frac{1}{x^3}={(x+\frac{1}{x})}^3-3(x+\frac{1}{x})$
Substituting $x+\frac{1}{x}=\sqrt{5}$:
$x^3+\frac{1}{x^3}={\left(\sqrt{5}\right)}^3-3\sqrt{5}=5\sqrt{5}-3\sqrt{5}=2\sqrt{5}$

Step 3: Determine the value of $x^6+\frac{1}{x^6}$
Squaring $x^3+\frac{1}{x^3}$:
${(x^3+\frac{1}{x^3})}^2=x^6+\frac{1}{x^6}+2$
Substituting $x^3+\frac{1}{x^3}=2\sqrt{5}$:
${\left(2\sqrt{5}\right)}^2=x^6+\frac{1}{x^6}+2$
$20=x^6+\frac{1}{x^6}+2$
$x^6+\frac{1}{x^6}=18$

Step 4: Compute the value of $x^9+\frac{1}{x^9}$
Consider the product of $(x^6+\frac{1}{x^6})$ and $(x^3+\frac{1}{x^3})$:
$(x^6+\frac{1}{x^6})(x^3+\frac{1}{x^3})=x^9+x^3+\frac{1}{x^3}+\frac{1}{x^9}$
Rearranging the terms:
$x^9+\frac{1}{x^9}=(x^6+\frac{1}{x^6})(x^3+\frac{1}{x^3})-(x^3+\frac{1}{x^3})$
Substituting the values we obtained:
$x^9+\frac{1}{x^9}=\left(18\right)\left(2\sqrt{5}\right)-2\sqrt{5}$
$x^9+\frac{1}{x^9}=36\sqrt{5}-2\sqrt{5}=34\sqrt{5}$