含立方根分式方程求解咨询：请求解指定代数方程

Hey there, let's work through this equation step by step. First, I notice you tried a substitution which is a great approach—but there's a small mistake in your algebraic manipulation that threw off the direction. Let's fix that and solve it properly.

First, let's recap your solid substitution choice:
Let $a = \sqrt[3]{34-x}$ and $b = \sqrt[3]{x+1}$. This gives us two key relationships:

• $a^3 = 34 - x$
• $b^3 = x + 1$

Correcting the Algebraic Simplification #

Looking at the numerator of the original equation: $(34-x)\sqrt[3]{x+1} - (x+1)\sqrt[3]{34-x} = a^3b - b^3a$. Your initial factoring was on the right track, but here's the critical correction:

$\frac{a^3b - b^3a}{a - b} = \frac{ab(a^2 - b^2)}{a - b}$
Since $a^2 - b^2 = (a - b)(a + b)$, we can cancel the $(a - b)$ terms (as long as $a \neq b$, which we'll check later) to get $ab(a + b)$, not $ab(a - b)$. That's the missing piece!

With this fix, the original equation simplifies to:
$$ab(a + b) = 30$$

Using the Sum of Cubes Identity #

Next, let's leverage another key detail from our substitution: add $a^3$ and $b^3$ together:
$$a^3 + b^3 = (34 - x) + (x + 1) = 35$$
We can use the sum of cubes identity to rewrite this:
$$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$
And we can rephrase $a^2 - ab + b^2$ as $(a + b)^2 - 3ab$ to make substitution easier. Let's define:

• $s = a + b$ (sum of our cube roots)
• $p = ab$ (product of our cube roots)

Now we have a system of two equations:

1. $p \cdot s = 30$ (from the corrected original equation)
2. $s(s^2 - 3p) = 35$ (from the sum of cubes)

Solving the System #

Substitute $p = \frac{30}{s}$ into the second equation:
$$s\left(s^2 - 3 \cdot \frac{30}{s}\right) = 35$$
Simplify the expression inside the parentheses:
$$s^3 - 90 = 35$$
$$s^3 = 125$$
Taking the real cube root gives $s = 5$.

Plug $s=5$ back into $p = \frac{30}{s}$ to get $p = 6$. Now we know:

• $a + b = 5$
• $ab = 6$

This means $a$ and $b$ are roots of the quadratic equation $t^2 - 5t + 6 = 0$. Factoring this gives $(t-2)(t-3)=0$, so $t=2$ or $t=3$.

Finding the Values of $x$ #

Case 1: $a=2$, $b=3$

• $\sqrt[3]{34 - x} = 2$ → $34 - x = 2^3 = 8$ → $x = 34 - 8 = 26$
• Verify with $b$: $\sqrt[3]{26 + 1} = \sqrt[3]{27} = 3$, which matches.

Case 2: $a=3$, $b=2$

• $\sqrt[3]{34 - x} = 3$ → $34 - x = 3^3 = 27$ → $x = 34 - 27 = 7$
• Verify with $b$: $\sqrt[3]{7 + 1} = \sqrt[3]{8} = 2$, which matches.

Checking for Invalid Solutions #

We need to ensure the original denominator $\sqrt[3]{34-x} - \sqrt[3]{x+1} \neq 0$. If $a = b$, then $34 - x = x + 1$ → $x = 16.5$, which isn't one of our solutions. So both $x=7$ and $x=26$ are valid.

内容的提问来源于stack exchange，提问作者G. Amber