Given:

\[ x = 1 + 2^{1/6} + 4^{1/6} + 8^{1/6} + 16^{1/6} + 32^{1/6} \]

Step 1: Write in powers of \( a = 2^{1/6} \)

\[ x = 1 + a + a^2 + a^3 + a^4 + a^5 = 1 + \frac{a(a^5 - 1)}{a - 1} \]

Step 2: Use identity \( a^6 = 2 \Rightarrow a^5 = \frac{2}{a} \)

\[ x = 1 + \frac{2 - a}{a - 1} = \frac{1}{a - 1} \Rightarrow 1 + \frac{1}{x} = a \Rightarrow \left(1 + \frac{1}{x} \right)^{24} = a^{24} \]

Step 3: Final calculation

\[ a = 2^{1/6} \Rightarrow a^{24} = (2^{1/6})^{24} = 2^4 = \boxed{16} \]

✅ Final Answer: $\boxed{16}$

Step 1: Recognize the series

The exponent is an infinite geometric series: $$ 1 + |\sin x| + |\sin x|^2 + |\sin x|^3 + \cdots $$

This is a geometric series with first term \( a = 1 \), common ratio \( r = |\sin x| \in [0,1] \), so: $$ \text{Sum} = \frac{1}{1 - |\sin x|} $$

Step 2: Rewrite the equation

$$ 5^{\frac{1}{1 - |\sin x|}} = 25 = 5^2 $$

Equating exponents: $$ \frac{1}{1 - |\sin x|} = 2 \Rightarrow 1 - |\sin x| = \frac{1}{2} \Rightarrow |\sin x| = \frac{1}{2} $$

Step 3: Solve for \( x \in (-\pi, \pi) \)

We want all \( x \in (-\pi, \pi) \) such that \( |\sin x| = \frac{1}{2} \)

So \( \sin x = \pm \frac{1}{2} \). Within \( (-\pi, \pi) \), the values of \( x \) satisfying this are:

• $x = \frac{\pi}{6}$
• $x = \frac{5\pi}{6}$
• $x = -\frac{\pi}{6}$
• $x = -\frac{5\pi}{6}$

✅ Final Answer: $\boxed{4}$ solutions

Problem:

If one Arithmetic Mean (AM) \( a \) and two Geometric Means \( p \) and \( q \) are inserted between any two positive numbers, find the value of: \[ p^3 + q^3 \]

Given:

• Let two positive numbers be \( A \) and \( B \).
• One AM: \( a = \frac{A + B}{2} \)
• Two GMs inserted: so the four terms in G.P. are: \[ A, \ p = \sqrt[3]{A^2B}, \ q = \sqrt[3]{AB^2}, \ B \]

Now calculate:

\[ pq = \sqrt[3]{A^2B} \cdot \sqrt[3]{AB^2} = \sqrt[3]{A^3B^3} = AB \]
\[ p^3 = A^2B, \quad q^3 = AB^2 \]
\[ p^3 + q^3 = A^2B + AB^2 = AB(A + B) \]

Also,

\[ 2apq = 2 \cdot \frac{A + B}{2} \cdot AB = AB(A + B) \]

✅ Therefore,

\( \boxed{p^3 + q^3 = 2apq} \)

? GP and Function Relation

Given: \( f(x) = x^3 + 3x - 9 \)

The sum of infinite GP = max value of \( f(x) \) on [−2, 3]

The difference between first two terms = \( f'(0) \)

Step 1: \( f(x) \) is increasing ⇒ Max at \( x = 3 \)

\( f(3) = 27 \Rightarrow \frac{a}{1 - r} = 27 \)

Step 2: \( f'(x) = 3x^2 + 3 \Rightarrow f'(0) = 3 \)

⇒ \( a(1 - r) = 3 \)

Step 3: Solve:
\( a = 27(1 - r) \)
\( \Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9} \Rightarrow r = \frac{2}{3} \)

✅ Final Answer: \( r = \frac{2}{3} \)