Sure, let's walk through the given problem step by step. The problem states that:
[tex]\[
x^2 + \frac{1}{x^2} = 34
\][/tex]
We need to find the value of:
[tex]\[
x^3 + \frac{1}{x^3} - 9
\][/tex]
First, we can analyze the given equation:
1. Solve the given equation [tex]\(x^2 + \frac{1}{x^2} = 34\)[/tex]:
- We start by letting [tex]\( u = x + \frac{1}{x} \)[/tex].
- From the given hint, we know:
[tex]\[
u^2 = x^2 + 2 + \frac{1}{x^2}
\][/tex]
- Rearranging the above equation:
[tex]\[
x^2 + \frac{1}{x^2} = u^2 - 2
\][/tex]
- From the given condition [tex]\( x^2 + \frac{1}{x^2} = 34 \)[/tex], we substitute:
[tex]\[
34 = u^2 - 2
\][/tex]
- Solving for [tex]\( u \)[/tex]:
[tex]\[
u^2 = 36 \implies u = \pm 6
\][/tex]
2. Determine the expression [tex]\( x^3 + \frac{1}{x^3} \)[/tex]:
- Notice that:
[tex]\[
(x + \frac{1}{x})^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}
\][/tex]
This can be rewritten as:
[tex]\[
(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})
\][/tex]
- Let [tex]\( v = x^3 + \frac{1}{x^3} \)[/tex], then:
[tex]\[
u^3 = v + 3u
\][/tex]
- Substitute [tex]\( u = 6 \)[/tex]:
[tex]\[
6^3 = v + 3 \cdot 6
\][/tex]
Solving for [tex]\( v \)[/tex]:
[tex]\[
216 = v + 18 \implies v = 198
\][/tex]
- Now, substitute [tex]\( u = -6 \)[/tex]:
[tex]\[
(-6)^3 = v + 3(-6)
\][/tex]
Solving for [tex]\( v \)[/tex]:
[tex]\[
-216 = v - 18 \implies v = -198
\][/tex]
3. Calculate [tex]\( x^3 + \frac{1}{x^3} - 9 \)[/tex]:
- For [tex]\( u = 6 \)[/tex]:
[tex]\[
x^3 + \frac{1}{x^3} = 198 \implies 198 - 9 = 189
\][/tex]
- For [tex]\( u = -6 \)[/tex]:
[tex]\[
x^3 + \frac{1}{x^3} = -198 \implies -198 - 9 = -207
\][/tex]
Therefore, the possible values for the donated amount are:
[tex]\[
₹189 \, \text{and} \, ₹-207
\][/tex]
Since donation cannot be negative, the negative value is not considered feasible.
Thus, Ashima donated ₹189 or -₹207 possible in steps, hence possible value ₹189 to the blind school.
