https://file.aiquickdraw.com/mathbot/file/1726821159058611r920r.jpg

Let's solve the equations given in the image one by one.

### (a) \( x^{\frac{2}{3}} + x^{\frac{1}{3}} - 6 = 0 \)

To simplify, let \( y = x^{\frac{1}{3}} \), which means \( y^2 = x^{\frac{2}{3}} \). The equation becomes:

\[
y^2 + y - 6 = 0
\]

This is a quadratic equation. Solving it using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 1 \), and \( c = -6 \):

\[
y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm \sqrt{25}}{2}
\]

\[
y = \frac{-1 \pm 5}{2}
\]

So, the two solutions for \( y \) are:

\[
y = \frac{-1 + 5}{2} = 2 \quad \text{or} \quad y = \frac{-1 - 5}{2} = -3
\]

Since \( y = x^{\frac{1}{3}} \), we now cube both sides to find \( x \):

- If \( y = 2 \), then \( x = 2^3 = 8 \).
- If \( y = -3 \), then \( x = (-3)^3 = -27 \).

Thus, the solutions for part (a) are:

\[
x = 8 \quad \text{or} \quad x = -27
\]

---

### (b) \( x^4 + 3x^2 - 10 = 0 \)

Let \( z = x^2 \), which means \( z^2 = x^4 \). The equation becomes:

\[
z^2 + 3z - 10 = 0
\]

This is also a quadratic equation. Solving it using the quadratic formula:

\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 3 \), and \( c = -10 \):

\[
z = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2}
\]

\[
z = \frac{-3 \pm 7}{2}
\]

So, the two solutions for \( z \) are:

\[
z = \frac{-3 + 7}{2} = 2 \quad \text{or} \quad z = \frac{-3 - 7}{2} = -5
\]

Since \( z = x^2 \), we now solve for \( x \):

- If \( z = 2 \), then \( x^2 = 2 \), so \( x = \pm \sqrt{2} \).
- If \( z = -5 \), there are no real solutions because \( x^2 = -5 \) has no real solutions.

Thus, the solutions for part (b) are:

\[
x = \pm \sqrt{2}
\]

---

### Final Answers:
(a) \( x = 8 \) or \( x = -27 \)  
(b) \( x = \pm \sqrt{2} \)

Would you like more details on any part of the solution or have any questions?

Here are five related questions to explore:
1. How do you factor quadratic equations when solving for variables?
2. What is the role of substitution in simplifying higher-order equations?
3. How can the quadratic formula be derived?
4. What are the conditions for a quadratic equation to have real solutions?
5. How would complex solutions appear in part (b)?

**Tip:** When solving higher-order equations, substitution methods can simplify them into more manageable forms like quadratics.

Solve these equations, giving your answers in exact form: (a) x^(2/3) + x^(1/3) - 6 = 0, (b) x^4 + 3x^2 - 10 = 0

Step-by-step guide on how to solve the algebraic equations x^(2/3) + x^(1/3) - 6 = 0 and x^4 + 3x^2 - 10 = 0 using substitution and the quadratic formula. Covers key algebraic concepts and theorems for students in Grades 10-12.

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