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.