This solution explains how to find the value of the expression \(8x^3-\frac{27}{x^3}\) given a specific quartic equation and a condition on the variable \(x\).

Solving the Quartic Equation for x

We are given the equation \(4x^4-37x^2+9=0\). This is a quartic equation, but it can be treated as a quadratic equation in terms of \(x^2\). Let's make a substitution: let \(y = x^2\). Substituting \(y\) into the equation gives us:

$$4(x^2)^2 - 37(x^2) + 9 = 0$$

$$4y^2 - 37y + 9 = 0$$

Now, we need to solve this quadratic equation for \(y\). We can solve it by factoring. We look for two numbers that multiply to \(4 \times 9 = 36\) and add up to \(-37\). These numbers are \(-36\) and \(-1\).

Rewrite the middle term:

$$4y^2 - 36y - y + 9 = 0$$

Factor by grouping:

$$4y(y - 9) - 1(y - 9) = 0$$

Group the common term \((y - 9)\):

$$(4y - 1)(y - 9) = 0$$

This gives us two possible values for \(y\):

$$4y - 1 = 0 \implies y = \frac{1}{4}$$

or

$$y - 9 = 0 \implies y = 9$$

Since we defined \(y = x^2\), we now have:

$$x^2 = \frac{1}{4} \quad \text{or} \quad x^2 = 9$$

Solving for \(x\) in each case:

If \(x^2 = \frac{1}{4}\), then \(x = \pm \sqrt{\frac{1}{4}}\), which means \(x = \frac{1}{2}\) or \(x = -\frac{1}{2}\).

If \(x^2 = 9\), then \(x = \pm \sqrt{9}\), which means \(x = 3\) or \(x = -3\).

So, the possible values for \(x\) are \(\frac{1}{2}, -\frac{1}{2}, 3, -3\).

Applying the Condition x > sqrt(3/2)

We are given the condition that \(x > \sqrt{\frac{3}{2}}\). Let's evaluate \(\sqrt{\frac{3}{2}}\) to understand its approximate value. \(\frac{3}{2} = 1.5\). We know that \(1^2 = 1\) and \(2^2 = 4\), so \(\sqrt{1.5}\) is a number between 1 and 2. Squaring the possible values of \(x\) and comparing them to \(\frac{3}{2}\) helps:

• For \(x = \frac{1}{2}\): \(x^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\). Since \(\frac{1}{4} = 0.25\) and \(\frac{3}{2} = 1.5\), \(0.25\) is not greater than \(1.5\). Thus, \(x = \frac{1}{2}\) does not satisfy \(x^2 > \frac{3}{2}\).
• For \(x = -\frac{1}{2}\): This value is negative, and \(\sqrt{\frac{3}{2}}\) is positive, so \(x = -\frac{1}{2}\) cannot be greater than \(\sqrt{\frac{3}{2}}\).
• For \(x = 3\): \(x^2 = 3^2 = 9\). Since \(9 > 1.5\), this value satisfies \(x^2 > \frac{3}{2}\). Therefore, \(x = 3\) is a valid solution.
• For \(x = -3\): This value is negative, and \(\sqrt{\frac{3}{2}}\) is positive, so \(x = -3\) cannot be greater than \(\sqrt{\frac{3}{2}}\).

Based on the condition \(x > \sqrt{\frac{3}{2}}\), the only value of \(x\) that satisfies the equation and the condition is \(x = 3\).

Evaluating the Expression 8x^3 - 27/x^3

Now we need to calculate the value of the expression \(8x^3-\frac{27}{x^3}\) using the value \(x = 3\).

Substitute \(x = 3\) into the expression:

$$8(3)^3 - \frac{27}{(3)^3}$$

Calculate the powers of 3:

$$3^3 = 3 \times 3 \times 3 = 27$$

Now substitute this value back into the expression:

$$8(27) - \frac{27}{27}$$

Perform the multiplication and division:

$$8 \times 27 = 216$$

$$\frac{27}{27} = 1$$

Finally, perform the subtraction:

$$216 - 1 = 215$$

Therefore, the value of the expression \(8x^3-\frac{27}{x^3}\) when \(x=3\) is 215.