The difference between the maximum real root and the minimum real root of the equation $(x^2 - 5)^4 + (x^2 - 7)^4 = 16$ is

The two bases $(x^2 - 5)$ and $(x^2 - 7)$ are centered around $6$, so let:

$u = x^2 - 6$

Then $x^2 - 5 = u + 1$ and $x^2 - 7 = u - 1$, and the equation becomes:

$(u + 1)^4 + (u - 1)^4 = 16$

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Expand:

$(u+1)^4 = u^4 + 4u^3 + 6u^2 + 4u + 1$

$(u-1)^4 = u^4 - 4u^3 + 6u^2 - 4u + 1$

Adding,

$2u^4 + 12u^2 + 2 = 16$

$u^4 + 6u^2 - 7 = 0$

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Let $v = u^2$:

$v^2 + 6v - 7 = 0$

$(v + 7)(v - 1) = 0$

So $v = 1$ or $v = -7$

Since $v = u^2 \geq 0$, we reject $v = -7$.

$u^2 = 1 \implies u = \pm 1$

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When $u = 1$: $x^2 = u + 6 = 7 \implies x = \pm\sqrt{7}$

When $u = -1$: $x^2 = u + 6 = 5 \implies x = \pm\sqrt{5}$

So the four real roots are: $-\sqrt{7},\ -\sqrt{5},\ \sqrt{5},\ \sqrt{7}$

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$\text{Max root} - \text{Min root} = \sqrt{7} - (-\sqrt{7}) = 2\sqrt{7}$