Let's solve the given equations from the uploaded image.

Problem 8:

$16x^3 = 54$

Steps to Solve:

1. Divide both sides by 16 to isolate $x^3$: $x^3 = \frac{54}{16} = \frac{27}{8}$

2. Take the cube root of both sides: [ x = \sqrt[3]{\frac{27}{8}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}} = \frac{3}{2} ]

Solution: $x = \frac{3}{2}$

Problem 10:

$x^4 + 3 = 13 - 9x^4$

Steps to Solve:

1. Combine like terms by adding $9x^4$ to both sides and subtracting 3: $x^4 + 9x^4 = 13 - 3$ $10x^4 = 10$

2. Divide both sides by 10: $x^4 = 1$

3. Solve for $x$ by taking the fourth root of both sides: $x = \pm\sqrt[4]{1} = \pm 1$

Solution: $x = \pm 1$

Summary of Solutions:

1. For Problem 8: $x = \frac{3}{2}$
2. For Problem 10: $x = \pm 1$

Would you like a detailed explanation for any of these steps?

Related Questions:

1. How do we simplify fractions like $\frac{54}{16}$?
2. What is the difference between cube roots and fourth roots?
3. How do we find the real and complex roots of $x^4 = 1$?
4. Why does the equation $x = \pm 1$ have two solutions?
5. Can these techniques be applied to more complex polynomial equations?

Tip:

Always simplify equations by isolating the variable first—this helps reduce mistakes during algebraic manipulations!