Let's break down the given expression step by step.

We are asked to find the value of $K$ in the expansion of:

$\left( x^7 \left( \frac{x^0 4}{2} + \frac{K}{x^3} \right) \right)^7$

Let's clarify the expression first. The original expression seems unclear with the notation provided. Based on interpretation, I assume the correct expression is:

$\left( x^7 \left( \frac{4}{2} + \frac{K}{x^3} \right) \right)^7$

Simplifying the inner terms:

$\left( x^7 \left( 2 + \frac{K}{x^3} \right) \right)^7$

Now, we distribute the $x^7$ inside the parentheses:

$\left( x^7 \cdot 2 + x^7 \cdot \frac{K}{x^3} \right)^7 = \left( 2x^7 + Kx^4 \right)^7$

We now expand $\left( 2x^7 + Kx^4 \right)^7$ using the binomial theorem:

$(2x^7 + Kx^4)^7 = \sum_{r=0}^{7} \binom{7}{r} (2x^7)^{7-r} (Kx^4)^r$

We want the constant term in the expansion. To find that, we look for the term where the powers of $x$ cancel out, i.e., where the exponent of $x$ is zero.

General term in the expansion:

$\binom{7}{r} (2x^7)^{7-r} (Kx^4)^r = \binom{7}{r} 2^{7-r} x^{7(7-r)} K^r x^{4r}$ $= \binom{7}{r} 2^{7-r} K^r x^{49 - 3r}$

For the constant term, the exponent of $x$ must be zero: $49 - 3r = 0$ Solving for $r$: $r = \frac{49}{3} = 7$

So, we need the term with $r = 7$.

Substituting $r = 7$ into the general term: $\binom{7}{7} 2^{7-7} K^7 x^{49 - 3(7)} = \binom{7}{7} K^7 x^0$

Thus, the constant term is $K^7$.

We are given that the constant term is 168, so: $K^7 = 168$

Taking the seventh root of both sides: $K = \sqrt[7]{168}$

Let me calculate the exact value of $K$.The value of $K$ is approximately $2.08$.

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