Question : If $x^4+\frac{1}{x^4}=194, x>0$, then find the value of $x^3+\frac{1}{x^3}+x+\frac{1}{x}$
Option 1: 76
Option 2: 66
Option 3: 56
Option 4: 46
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Correct Answer: 56
Solution :
Given: $x^4+\frac{1}{x^4}=194$
$⇒x^4+\frac{1}{x^4}+2=194+2$
$⇒(x^2+\frac{1}{x^2})^2=196$
$⇒x^2+\frac{1}{x^2}=\sqrt{196}$
$⇒x^2+\frac{1}{x^2}=14$
$⇒x^2+\frac{1}{x^2}+2=14+2$
$⇒(x+\frac{1}{x})^2=16$
$⇒x+\frac{1}{x}=4$ -----------------------------(1)
Now, $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3×x×\frac{1}{x}(x+\frac{1}{x})$
$⇒x^3+\frac{1}{x^3}=4^3-3 ×4$
$⇒x^3+\frac{1}{x^3}=52$--------------------------------(2)
Therefore, $x^3+\frac{1}{x^3}+x+\frac{1}{x}$ = $52+4 = 56$ (from equations (1) and (2))
Hence, the correct answer is 56.
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