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Sagot :
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.
[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.
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