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प्रश्न
If x2 – 6x + 1 = 0, find the value of `x^3 + 1/x^3`.
बेरीज
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उत्तर
Given: x2 – 6x + 1 = 0
Step-wise calculation:
1. Divide the equation by (x) assuming (x ≠ 0):
`x - 6 + 1/x = 0`
⇒ `x + 1/x = 6`
2. Square both sides to find `x^2 + 1/x^2`:
`(x + 1/x)^2 = 6^2`
⇒ `x^2 + 2 + 1/x^2 = 36`
`x^2 + 1/x^2 = 36 - 2`
`x^2 + 1/x^2 = 34`
3. Use the identity to find `x^3 + 1/x^3`:
`(x + 1/x)^3 = x^3 + 1/x^3 + 3(x + 1/x)`
Substitute `x + 1/x = 6`:
`6^3 = x^3 + 1/x^3 + 3 xx 6`
`216 = x^3 + 1/x^3 + 18`
`x^3 + 1/x^3 = 216 - 18`
`x^3 + 1/x^3 = 198`
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पाठ 3: Expansions - Exercise 3B [पृष्ठ ७२]
