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Let y = x + \frac{1}{x}. Then, squaring both sides:

y^2 = x^2 + 2 + \frac{1}{x^2} \Rightarrow x^2 + \frac{1}{x^2} = y^2 - 2

Now square again:

(x^2 + \frac{1}{x^2})^2 = x^4 + 2 + \frac{1}{x^4} \Rightarrow x^4 + \frac{1}{x^4} = (y^2 - 2)^2 - 2

Given x^4 + \frac{1}{x^4} = 47:

(y^2 - 2)^2 - 2 = 47

Solve for y^2:

(y^2 - 2)^2 = 49 \Rightarrow y^2 - 2 = \pm 7

So y^2 = 9 or y^2 = -5. Since y^2 \geq 0, y^2 = 9 \Rightarrow y = \pm 3

Now, x^3 + \frac{1}{x^3} = y^3 - 3y

If y = 3: 3^3 - 3(3) = 27 - 9 = 18

If y = -3: (-3)^3 - 3(-3) = -27 + 9 = -18

Thus, x^3 + \frac{1}{x^3} = \pm 18

हिन्दी उत्तर:

माना y = x + \frac{1}{x}. तब दोनों तरफ वर्ग करने पर:

y^2 = x^2 + 2 + \frac{1}{x^2} \Rightarrow x^2 + \frac{1}{x^2} = y^2 - 2

अब फिर से वर्ग करें:

(x^2 + \frac{1}{x^2})^2 = x^4 + 2 + \frac{1}{x^4} \Rightarrow x^4 + \frac{1}{x^4} = (y^2 - 2)^2 - 2

दिया गया है x^4 + \frac{1}{x^4} = 47:

(y^2 - 2)^2 - 2 = 47

y^2 के लिए हल करें:

(y^2 - 2)^2 = 49 \Rightarrow y^2 - 2 = \pm 7

तो y^2 = 9 या y^2 = -5. चूँकि y^2 \geq 0, इसलिए y^2 = 9 \Rightarrow y = \pm 3

अब, x^3 + \frac{1}{x^3} = y^3 - 3y

यदि y = 3: 3^3 - 3(3) = 27 - 9 = 18

यदि y = -3: (-3)^3 - 3(-3) = -27 + 9 = -18

इसलिए, x^3 + \frac{1}{x^3} = \pm 18