Verified by Ajay Kumar
If x + 1/x = 9, then find the value of x³ + 1/x³ + x² + 1/x².
702
729
779
781
Answer: D
Solution: Given \(x+\frac1x=9\). First, \(x^2+\frac1{x^2}=9^2-2=79\). Next, \(x^3+\frac1{x^3}=(x+\frac1x)^3-3(x+\frac1x)=9^3-3\times9=702\). The required sum \(=79+702=781\). Trick: Use the square identity first, then the cube identity, and add the two results.
यदि x + 1/x = 9 है, तो x³ + 1/x³ + x² + 1/x² का मान ज्ञात कीजिए।
702
729
779
781
उत्तर: D
हल: दिया है \(x+\frac1x=9\)। पहले \(x^2+\frac1{x^2}=9^2-2=79\) निकालें। फिर \(x^3+\frac1{x^3}=(x+\frac1x)^3-3(x+\frac1x)=9^3-3\times9=702\)। आवश्यक योग \(=79+702=781\)। ट्रिक: पहले वर्ग वाली सर्वसमिका, फिर घन वाली सर्वसमिका लगाएँ और दोनों परिणाम जोड़ दें।
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