The problem provides the equation $x^2 = 5 + 2\sqrt{6}$ and asks to find the value of $x$. Also, it states that $a+b+c=m$, $a^2+b^2+c^2=n$ and $a^3+b^3 = p^3$. We need to show that $\frac{x^8+1}{x^4} = 98$, and if $c=0$, we need to show that $m^3 + 2p^3 = 3mn$.

1. Problem Description

The problem provides the equation $x^2 = 5 + 2\sqrt{6}$ and asks to find the value of $x$. Also, it states that $a+b+c=m$, $a^2+b^2+c^2=n$ and $a^3+b^3 = p^3$. We need to show that $\frac{x^8+1}{x^4} = 98$, and if $c=0$, we need to show that $m^3 + 2p^3 = 3mn$.

2. Solution Steps

Part (a): Find the value of $x$.

We are given $x^2 = 5 + 2\sqrt{6}$. We can rewrite the right side as a perfect square:

$x^2 = 5 + 2\sqrt{6} = 2 + 3 + 2\sqrt{2}\sqrt{3} = (\sqrt{2})^2 + (\sqrt{3})^2 + 2\sqrt{2}\sqrt{3} = (\sqrt{2} + \sqrt{3})^2$

Taking the square root of both sides, we get

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

Since the question asks for "the value of x", we'll take the positive root.

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

Part (b): Show that $\frac{x^8+1}{x^4} = 98$.

Since $x = \sqrt{2} + \sqrt{3}$, we have $x^2 = 5 + 2\sqrt{6}$.

We want to find $x^4$.

$x^4 = (x^2)^2 = (5 + 2\sqrt{6})^2 = 5^2 + 2(5)(2\sqrt{6}) + (2\sqrt{6})^2 = 25 + 20\sqrt{6} + 4(6) = 25 + 20\sqrt{6} + 24 = 49 + 20\sqrt{6}$.

Now we want to find $x^8$.

$x^8 = (x^4)^2 = (49 + 20\sqrt{6})^2 = 49^2 + 2(49)(20\sqrt{6}) + (20\sqrt{6})^2 = 2401 + 1960\sqrt{6} + 400(6) = 2401 + 1960\sqrt{6} + 2400 = 4801 + 1960\sqrt{6}$.

Now we consider $\frac{x^8+1}{x^4} = \frac{4801 + 1960\sqrt{6} + 1}{49 + 20\sqrt{6}} = \frac{4802 + 1960\sqrt{6}}{49 + 20\sqrt{6}}$.

We can factor out 98 from the numerator:

$4802 = 98 \times 49$ and $1960 = 98 \times 20$.

So, $\frac{x^8+1}{x^4} = \frac{98(49 + 20\sqrt{6})}{49 + 20\sqrt{6}} = 98$.

Part (c): If $c=0$, show that $m^3 + 2p^3 = 3mn$.

We are given $a+b+c=m$, $a^2+b^2+c^2=n$, and $a^3+b^3=p^3$. If $c=0$, then $m = a+b$ and $n = a^2+b^2$. We want to show that $m^3 + 2p^3 = 3mn$.

$m^3 = (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a+b) = p^3 + 3abm$

$m^3 + 2p^3 = p^3 + 3abm + 2p^3 = 3p^3 + 3abm = 3(a^3+b^3) + 3ab(a+b)$

Also, $3mn = 3(a+b)(a^2+b^2) = 3(a^3 + ab^2 + a^2b + b^3) = 3(a^3+b^3 + ab(a+b))$.

Therefore, $3mn = 3(a^3+b^3) + 3ab(a+b) = 3p^3 + 3abm$.

So, $m^3 + 2p^3 = 3p^3 + 3abm = 3mn$.

Hence, $m^3 + 2p^3 = 3mn$ when $c=0$.

3. Final Answer

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

(b) $\frac{x^8+1}{x^4} = 98$

(c) $m^3 + 2p^3 = 3mn$