$\left( vii \right)\text{ }Given,\text{ }{{\left( x~+\text{ }2 \right)}^{3}}~=\text{ }2x({{x}^{2}}~\text{ }1)$

By using the formula for ${{\left( a+b \right)}^{2~}}=\text{ }{{a}^{2}}+2ab+{{b}^{2}}$

$\Rightarrow ~{{x}^{3}}~+\text{ }8~+~{{x}^{2}}~+\text{ }12x~=\text{ }2{{x}^{3}}~\text{ }2x$

$\Rightarrow ~{{x}^{3}}~+\text{ }14x~\text{ }6{{x}^{2}}~\text{ }8\text{ }=\text{ }0$

Since the above equation is not in the form of $a{{x}^{2}}~+~bx~+~c~=\text{ }0$ .

Therefore, the given equation is not a quadratic equation.

$\left( viii \right)\text{ }Given,~{{x}^{3}}~\text{ }4{{x}^{2}}~~x~+\text{ }1\text{ }=\text{ }{{\left( x~\text{ }2 \right)}^{3}}$

By using the formula for ${{\left( a+b \right)}^{2~}}=\text{ }{{a}^{2}}+2ab+{{b}^{2}}$

$\Rightarrow ~{{x}^{3}}~\text{ }4{{x}^{2}}~~x~+\text{ }1~=~{{x}^{3}}~\text{ }8\text{ }\text{ }6{{x}^{2~}}~+\text{ }12x$

$\Rightarrow 2{{x}^{2}}~\text{ }13x~+\text{ }9\text{ }=\text{ }0$

Since the above equation is in the form of $a{{x}^{2}}~+~bx~+~c~=\text{ }0.$

Therefore, the given equation is quadratic equation.