To solve the equation \(8x^6 + 215x^3 - 27 = 0\), we can use substitution. Let \(y = x^3\). Then the equation becomes:

\(8y^2 + 215y - 27 = 0\)

We can factor this quadratic equation as:

\((8y - 1)(y + 27) = 0\)

Setting each factor to zero gives:

\(8y - 1 = 0 \quad \Rightarrow \quad y = \frac{1}{8}\)

\(y + 27 = 0 \quad \Rightarrow \quad y = -27\)

Since \(y = x^3\), we have:

\(x^3 = \frac{1}{8} \quad \Rightarrow \quad x = \frac{1}{2}\)

\(x^3 = -27 \quad \Rightarrow \quad x = -3\)

Thus, the solutions are \(x = \frac{1}{2}\) and \(x = -3\).

Let \(u = 2x - 3\). Then \(u^2 = (2x - 3)^2\).

Substitute into the equation:

\(u^2 - \frac{4}{u^2} - 3 = 0\).

Multiply through by \(u^2\) to eliminate the fraction:

\(u^4 - 3u^2 - 4 = 0\).

Let \(v = u^2\), then:

\(v^2 - 3v - 4 = 0\).

Factorize:

\((v - 4)(v + 1) = 0\).

So, \(v = 4\) or \(v = -1\).

Since \(v = u^2\), \(u^2 = 4\) or \(u^2 = -1\).

\(u^2 = -1\) has no real solutions.

For \(u^2 = 4\), \(u = \pm 2\).

Substitute back for \(u = 2x - 3\):

\(2x - 3 = 2\) or \(2x - 3 = -2\).

Solving these gives:

\(2x = 5\) or \(2x = 1\).

Thus, \(x = \frac{5}{2}\) or \(x = \frac{1}{2}\).

Let \(u = x^{\frac{1}{2}}\), then \(u^2 = x\).

The equation becomes:

\(4u^2 - 11u + 6 = 0\)

Factor the quadratic equation:

\((4u - 3)(u - 2) = 0\)

Set each factor to zero:

\(4u - 3 = 0 \quad \text{or} \quad u - 2 = 0\)

Solve for \(u\):

\(u = \frac{3}{4} \quad \text{or} \quad u = 2\)

Since \(u = x^{\frac{1}{2}}\), square both sides to find \(x\):

\(x = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \quad \text{or} \quad x = 2^2 = 4\)

Thus, the solutions are \(x = \frac{9}{16}\) or \(x = 4\).

To find the intersection points, set the equations equal to each other:

\(x^{\frac{2}{3}} - 1 = x^{\frac{1}{3}} + 1\).

Rearrange to form:

\(x^{\frac{2}{3}} - x^{\frac{1}{3}} - 2 = 0\).

Let \(u = x^{\frac{1}{3}}\), then \(u^2 = x^{\frac{2}{3}}\).

The equation becomes:

\(u^2 - u - 2 = 0\).

Factor the quadratic:

\((u - 2)(u + 1) = 0\).

Thus, \(u = 2\) or \(u = -1\).

Convert back to \(x\):

For \(u = 2\), \(x = 2^3 = 8\).

For \(u = -1\), \(x = (-1)^3 = -1\).

Find \(y\) for each \(x\):

For \(x = 8\), \(y = 8^{\frac{1}{3}} + 1 = 2 + 1 = 3\).

For \(x = -1\), \(y = (-1)^{\frac{1}{3}} + 1 = -1 + 1 = 0\).

Thus, the points of intersection are \((8, 3)\) and \((-1, 0)\).

Let \(y = \frac{1}{x^2}\). Then \(y^2 = \frac{1}{x^4}\).

Substitute into the equation:

\(18y^2 + y = 4\)

Rearrange to form a quadratic equation:

\(18y^2 + y - 4 = 0\)

Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 18\), \(b = 1\), \(c = -4\):

\(y = \frac{-1 \pm \sqrt{1^2 - 4 \times 18 \times (-4)}}{2 \times 18}\)

\(y = \frac{-1 \pm \sqrt{1 + 288}}{36}\)

\(y = \frac{-1 \pm \sqrt{289}}{36}\)

\(y = \frac{-1 \pm 17}{36}\)

\(y = \frac{16}{36} \text{ or } y = \frac{-18}{36}\)

\(y = \frac{4}{9} \text{ or } y = -\frac{1}{2}\)

Since \(y = \frac{1}{x^2}\), solve for \(x\):

For \(y = \frac{4}{9}\):

\(\frac{1}{x^2} = \frac{4}{9}\)

\(x^2 = \frac{9}{4}\)

\(x = \pm \frac{3}{2}\)

For \(y = -\frac{1}{2}\), \(x^2\) cannot be negative, so ignore.

Thus, the real roots are \(x = 1.5\) and \(x = -1.5\).