Question : If $x^{2}+\frac{1}{x^{2}} = 98(x>0)$, then the value of $x^{3}+\frac{1}{x^{3}}$ is:
Option 1: 970
Option 2: 1030
Option 3: –970
Option 4: –1030
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Correct Answer: 970
Solution : Given: $x^{2}+\frac{1}{x^{2}} = 98$ We know, $(x+y)^{2}= x^{2}+y^{2}+2xy$ ⇒ $(x+\frac{1}{x})^{2}= x^{2}+\frac{1}{x^{2}}+2(x)(\frac{1}{x})$ ⇒ $(x+\frac{1}{x})^{2}= 98+2$ ⇒ $(x+\frac{1}{x})^{2}= 100$ ⇒ $ (x+\frac{1}{x})= 10$ Now, $x^{3}+\frac{1}{x^{3}} = (x+\frac{1}{x})^{3}-3(x)(\frac{1}{x})(x+\frac{1}{x})$ = $10^{3}-3(10)$ = $1000-30$ = $970$ Hence, the correct answer is 970.
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