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

Question

Question : If $\frac{x^{2} + 1}{x} = 5$ , then find the value of $x^{4} + \frac{1}{x^{4}} - 36$ .

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Answer 1

Correct Answer: 56

Solution : Given: $x^{4} + \frac{1}{x^{4}} = 194$
$\Rightarrow x^{4} + \frac{1}{x^{4}} + 2 = 194 + 2$
$\Rightarrow (x^{2} + \frac{1}{x^{2}})^{2} = 196$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = \sqrt{196}$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 14$
$\Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 14 + 2$
$\Rightarrow (x + \frac{1}{x})^{2} = 16$
$\Rightarrow x + \frac{1}{x} = 4$ -----------------------------(1)
Now, $x^{3} + \frac{1}{x^{3}} = (x + \frac{1}{x})^{3} - 3 \times x \times \frac{1}{x} (x + \frac{1}{x})$
$\Rightarrow x^{3} + \frac{1}{x^{3}} = 4^{3} - 3 \times 4$
$\Rightarrow x^{3} + \frac{1}{x^{3}} = 52$ --------------------------------(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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Question : If x4+1x4=194,x>0, then find the value of x3+1x3+x+1xOption 1: 76Option 2: 66Option 3: 56Option 4: 46

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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