如何在不计算t值的情况下求解该有理三次方程的无理式根？

Great job working through the substitution method for this cubic equation—your derivation is totally correct using the core logic of Cardano's method. Let’s break down how to simplify that messy-looking radical expression without crunching any decimal values:

Step 1: Find the rational root of the original equation #

For cubic equations with integer coefficients, the Rational Root Theorem tells us any possible rational root is a factor of the constant term divided by a factor of the leading coefficient. Here, the constant term is -18 and leading coefficient is 1, so possible roots are ±1, ±2, ±3, ±6, ±9, ±18.

Test $x=2$:
$$2^3 + 5*2 - 18 = 8 + 10 - 18 = 0$$
Perfect! $x=2$ is a real root of the equation.

Step 2: Connect your radical expression to this root #

You derived the expression:
$$x = \sqrt[3]{9+\sqrt{81+\frac{125}{27}}} + \sqrt[3]{9-\sqrt{81+\frac{125}{27}}}$$

Let’s define:

• $A = \sqrt[3]{9+\sqrt{81+\frac{125}{27}}}$
• $B = \sqrt[3]{9-\sqrt{81+\frac{125}{27}}}$

From your earlier steps, we already know two key facts:

1. $A^3 + B^3 = 18$ (this came from setting $3ab+5=0$, which eliminated the $(a+b)$ term)
2. $A^3B3 = (ab)^3 = \left(-\frac{5}{3}\right)^3 = -\frac{125}{27}$, so $AB = \sqrt[3]{-\frac{125}{27}} = -\frac{5}{3}$

Now expand $x^3 = (A+B)^3$ using the cube-of-a-sum formula:
$$x^3 = A^3 + B^3 + 3AB(A+B)$$
Substitute the values we have:
$$x^3 = 18 + 3*\left(-\frac{5}{3}\right)*x$$
Simplify to get back your original equation:
$$x^3 = 18 - 5x \implies x^3 + 5x - 18 = 0$$

For this cubic equation, the discriminant (calculated as $\Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3$ for $x^3+px+q=0$) is positive:
$$\Delta = \left(-9\right)^2 + \left(\frac{5}{3}\right)^3 = 81 + \frac{125}{27} > 0$$
This means the equation has exactly one real root and two complex conjugate roots. Your radical expression is a real number (the first cube root is positive, the second is negative, and their sum is real), so it must equal the only real root we found: 2.

Step 3: Explicitly prove the sum equals 2 #

If you want to directly confirm the radical sum is 2, assume $2 = \sqrt[3]{m} + \sqrt[3]{n}$. Cube both sides:
$$8 = m + n + 3\sqrt[3]{mn}*2$$
We know from your derivation that $m + n = 18$, so substitute that in:
$$8 = 18 + 6\sqrt[3]{mn}$$
Solve for $\sqrt[3]{mn}$:
$$6\sqrt[3]{mn} = 8 - 18 = -10 \implies \sqrt[3]{mn} = -\frac{5}{3}$$
Cube both sides to get $mn = -\frac{125}{27}$, which exactly matches $mn = 81 - \left(81+\frac{125}{27}\right) = -\frac{125}{27}$ for your $m = 9+\sqrt{81+\frac{125}{27}}$ and $n = 9-\sqrt{81+\frac{125}{27}}$. This confirms the sum is indeed 2.

内容的提问来源于stack exchange，提问作者user126456