Question : If $x^4+\frac{1}{x^4}=1154$, where $x>0$, then what is the value of $x^3+\frac{1}{x^3}?$
Option 1: 205
Option 2: 214
Option 3: 185
Option 4: 198
Correct Answer: 198
Solution :
Given,
$x^4+\frac{1}{x^4}=1154$
Adding 2 on both sides,
$x^4+\frac1{x^4}+2=1154+2$
⇒ $(x^2)^2+(\frac1{x^2})^2+2\times x^2\times\frac1{x^2}=1156$
We know, $a^2+b^2+2ab=(a+b)^2$
⇒ $(x^2+\frac{1}{x^2})^2=1156$
⇒ $x^2+\frac{1}{x^2}=\sqrt{1156}$
⇒ $x^2+\frac{1}{x^2}=34$
Again, adding 2 on both sides,
⇒ $x^2+\frac{1}{x^2}+2=34+2$
⇒ $(x+\frac{1}{x})^2=36$
⇒ $x+\frac{1}{x}=6$
We know, $a^3+b^3=(a+b)(a^2-ab+b^2)$
⇒ $x^3+\frac{1}{x^3}=(x+\frac1x)(x^2-x\times\frac1x+\frac1{x^2})$
$= (x+\frac1x)(x^2+\frac1{x^2}-1)$
$= 6 × (34 – 1) = 6 × 33 = 198$
Hence, the correct answer is 198.
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