If $x=\sqrt{2}+\sqrt{3}$, evaluate the value of the algebraic expression $x^3-3x$.

The correct answer is $8\sqrt{2}+6\sqrt{3}$.

Step 1: Understand the given information

We are given the value of $x$:

$x=\sqrt{2}+\sqrt{3}$

We need to evaluate the value of the algebraic expression:

$x^3-3x$

Step 2: Calculate $x^3$

Using the algebraic identity ${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3$, set $a=\sqrt{2}$ and $b=\sqrt{3}$:

$x^3={(\sqrt{2}+\sqrt{3})}^3$

$x^3={\left(\sqrt{2}\right)}^3+3{\left(\sqrt{2}\right)}^2\left(\sqrt{3}\right)+3\left(\sqrt{2}\right){\left(\sqrt{3}\right)}^2+{\left(\sqrt{3}\right)}^3$

Simplify each term individually:

${\left(\sqrt{2}\right)}^3=2\sqrt{2}$

$3{\left(\sqrt{2}\right)}^2\left(\sqrt{3}\right)=3\times 2\times \sqrt{3}=6\sqrt{3}$

$3\left(\sqrt{2}\right){\left(\sqrt{3}\right)}^2=3\times \sqrt{2}\times 3=9\sqrt{2}$

${\left(\sqrt{3}\right)}^3=3\sqrt{3}$

Substitute these back into the expression for $x^3$:

$x^3=2\sqrt{2}+6\sqrt{3}+9\sqrt{2}+3\sqrt{3}$

Combine like terms:

$x^3=(2\sqrt{2}+9\sqrt{2})+(6\sqrt{3}+3\sqrt{3})$

$x^3=11\sqrt{2}+9\sqrt{3}$

Step 3: Calculate $3x$

Multiply $x$ by $3$:

$3x=3(\sqrt{2}+\sqrt{3})$

$3x=3\sqrt{2}+3\sqrt{3}$

Step 4: Evaluate $x^3-3x$

Subtract $3x$ from $x^3$:

$x^3-3x=(11\sqrt{2}+9\sqrt{3})-(3\sqrt{2}+3\sqrt{3})$

$x^3-3x=11\sqrt{2}+9\sqrt{3}-3\sqrt{2}-3\sqrt{3}$

Group terms with like radicals:

$x^3-3x=(11\sqrt{2}-3\sqrt{2})+(9\sqrt{3}-3\sqrt{3})$

$x^3-3x=8\sqrt{2}+6\sqrt{3}$

Hence, the evaluated value of the algebraic expression is $8\sqrt{2}+6\sqrt{3}$.