$$ 4 x^{3}=27 \quad $$ b, $$ 9 x^{3}=72 \quad $$ c, $$ \frac{1}{3} x^{5}=81 $$ d $$ 2 x^{3}-4=2 x 6 $$

Question

UpStudy AI · Accepted Answer

Sure, let's solve each of the given equations step by step.

---

### **Problem b**

$$
4x^{3} = 27
$$

**Solution:**

1. **Isolate $$ x^3 $$:**
   $$
   x^{3} = \frac{27}{4}
   $$
   
2. **Take the cube root of both sides:**
   $$
   x = \sqrt[3]{\frac{27}{4}} = \frac{\sqrt[3]{27}}{\sqrt[3]{4}} = \frac{3}{\sqrt[3]{4}}
   $$
   
   Alternatively, you can write:
   $$
   x = 3 \cdot 4^{-\frac{1}{3}}
   $$

---

### **Problem c**

$$
9x^{3} = 72
$$

**Solution:**

1. **Isolate $$ x^3 $$:**
   $$
   x^{3} = \frac{72}{9} = 8
   $$
   
2. **Take the cube root of both sides:**
   $$
   x = \sqrt[3]{8} = 2
   $$

---

### **Problem d**

This problem appears to include two separate equations:

1. $$\frac{1}{3}x^{5} = 81$$
2. $$2x^{3} - 4 = 2x + 6$$

Let's solve each one individually.

---

#### **Equation 1:**

$$
\frac{1}{3}x^{5} = 81
$$

**Solution:**

1. **Multiply both sides by 3 to isolate $$ x^5 $$:**
   $$
   x^{5} = 81 \times 3 = 243
   $$
   
2. **Take the fifth root of both sides:**
   $$
   x = \sqrt[5]{243}
   $$
   
   Since $$ 243 = 3^5 $$:
   $$
   x = 3
   $$

---

#### **Equation 2:**

$$
2x^{3} - 4 = 2x + 6
$$

**Solution:**

1. **Rearrange the equation to set it to zero:**
   $$
   2x^{3} - 2x - 10 = 0
   $$
   
2. **Simplify by dividing the entire equation by 2:**
   $$
   x^{3} - x - 5 = 0
   $$
   
3. **Solve for $$ x $$ using numerical methods (since it's a cubic equation without easily factorable roots):**

- **Initial Guess:** Let’s try $$ x = 2 $$:
     $$
     f(2) = 2^3 - 2 - 5 = 8 - 2 - 5 = 1 \quad (\text{Too high})
     $$
   
   - **Next Guess:** Try $$ x = 1.9 $$:
     $$
     f(1.9) = 1.9^3 - 1.9 - 5 \approx 6.859 - 1.9 - 5 = -0.041
     $$
     
   - **Refinement:** Using methods like Newton-Raphson, we find:
     $$
     x \approx 1.903
     $$
   
   **Thus, the approximate real solution is:**
   $$
   x \approx 1.903
   $$

---

### **Summary of Solutions**

- **Problem b:** $$ x = \dfrac{3}{\sqrt[3]{4}} $$ or $$ x = 3 \cdot 4^{-\frac{1}{3}} $$
- **Problem c:** $$ x = 2 $$
- **Problem d:**
  - First equation: $$ x = 3 $$
  - Second equation: $$ x \approx 1.903 $$

If you need further clarification or assistance with any of these problems, feel free to ask!
Solutions: - **Problem b:** $$ x = \frac{3}{\sqrt[3]{4}} $$ or $$ x = 3 \cdot 4^{-\frac{1}{3}} $$ - **Problem c:** $$ x = 2 $$ - **Problem d:** - First equation: $$ x = 3 $$ - Second equation: $$ x \approx 1.903 $$

\( 4 x^{3}=27 \quad \) b, \( 9 x^{3}=72 \quad \) c, \( \frac{1}{3} x^{5}=81 \) d \( 2 x^{3}-4=2 x 6 \)

Real Tutor Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Algebra Questions