Tìm x 4x^(-3) + 4x^(-6) = -68

To solve the equation \(4x^{-3} + 4x^{-6} = -68\), let's start by substituting variables for simplicity.

Let \(y = x^{-3}\). Then, \(x^{-6} = (x^{-3})^2 = y^2\). The equation now becomes:

\[
4y + 4y^2 = -68
\]

Next, we can rewrite the equation:

\[
4y^2 + 4y + 68 = 0
\]

To simplify, divide the entire equation by 4:

\[
y^2 + y + 17 = 0
\]

Now, we can use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 1\), and \(c = 17\):

\[
y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 17}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 68}}{2} = \frac{-1 \pm \sqrt{-67}}{2}
\]

Since the discriminant is negative, this means there are no real solutions for \(y\). Specifically:

\[
y = \frac{-1 \pm i\sqrt{67}}{2}
\]

Substituting back for \(x\), we find that \(x^{-3} = y\), leading to complex solutions. Therefore, the equation does not yield real values for \(x\).

In conclusion, the equation \(4x^{-3} + 4x^{-6} = -68\) has no real solutions.