Question : If $x^{4}+\frac{1}{x^{4}}=34$, what is the value of $x^{3}-\frac{1}{x^{3}} $?
Option 1: 0
Option 2: 6
Option 3: 8
Option 4: 14
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Correct Answer: 14
Solution :
Given: $x^{4}+\frac{1}{x^{4}}=34$
Adding 2 to both sides, we get,
$⇒x^{4}+\frac{1}{x^{4}}+2=34+2$
$⇒(x^{2}+\frac{1}{x^{2}})^{2}=(6)^{2}$
$⇒x^{2}+\frac{1}{x^{2}}=6$
Subtracting 2 from both sides, we get,
$⇒x^{2}+\frac{1}{x^{2}}-2=6-2$
$⇒(x-\frac{1}{x})^{2}=(2)^{2}$
$⇒x-\frac{1}{x}=2$
Now, $(x-\frac{1}{x})^3=x^3-\frac{1}{x^3}-3×x×\frac{1}{x}(x-\frac{1}{x})$
$⇒2^3=x^3-\frac{1}{x^3}-3×2$
$\therefore x^{3}-\frac{1}{x^{3}}=14$
Hence, the correct answer is 14.
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