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Question

If $$\displaystyle { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } =194$$, find the value of $$\displaystyle { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } $$.


A
64
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B
52
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C
40
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D
38
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Solution

The correct option is C 52
Given $$ : x^{4}+\dfrac{1}{x^{4}} = 194 $$
To find $$ : x^{3}+\dfrac{1}{x^{3}} = ? $$
$$ x^{4}+\dfrac{1}{x^{4}} = 194 $$
$$ (x^{2})^{2}+\left ( \dfrac{1}{x^{2}} \right )^{2} = 194 $$
$$ \left ( x^{2}+\dfrac{1}{x^{2}} \right )^{2} = 194 $$
$$ \left ( x^{2}+\dfrac{1}{x^2} \right )^{2}-2x^{2}\dfrac{1}{x^{2}} = 194 $$    $$(a^{2}+b^{2} = (a+b)^{2}-2ab) $$
$$ \left ( x^{2}+\dfrac{1}{x^{2}} \right )^{2} = 194+2 $$
$$ x^{2}+\dfrac{1}{x^{2}} = \sqrt{196} $$   $$ \boxed {x^{2}+\dfrac{1}{x^{2}} = 14}...(1) $$
$$ \left ( x+\dfrac{1}{x} \right ) ^{2}-2.x\dfrac{1}{x} = 14 $$
$$ [a^{2}+b^{2} = (a+b)^{2}-2ab] $$
$$ \left ( x+\dfrac{1}{x} \right )^{2} = 16 $$
$$ \boxed {x +\dfrac{1}{x} = 4}...(2) $$
$$ \therefore x^{3}+\dfrac{1}{x^{3}} = \left ( x+\dfrac{1}{x} \right )\left ( x^{2}+\dfrac{1}{x^{2}}-1 \right ) $$
from (1) & (2)
$$ x^{3}+\dfrac{1}{x^{3}} = 4\times (14-1) $$
$$ = 4\times 13 $$
$$ = \boxed {52} $$ 

1188443_316701_ans_8b04f3dc3b89435b96b5214e2d9f4755.jpg

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