Question

If $x+\frac{1}{x}=3$, then what is the value of $x^8+\frac{1}{x^8}?$

• A. 2207
• B. 1097
• C. 3407
• D. 47

Answer 1

The correct option is A 2207

Let's use identity
$(a+b{)}^2=a^2+2ab+b^2$.

$x+\frac{1}{x}=3$
On squaring both sides, we get:
${(x+\frac{1}{x})}^2=3^2$
$x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=9$
$x^2+2+\frac{1}{x^2}=9$
$x^2+\frac{1}{x^2}=9-2$
$x^2+\frac{1}{x^2}=7$

On squaring both sides, we get:
${(x^2+\frac{1}{x^2})}^2=7^2$
$x^4+2\times x^2\times \frac{1}{x^2}+{\left(\frac{1}{x^2}\right)}^2=49$
$x^4+2+\frac{1}{x^4}=49$
$x^4+\frac{1}{x^4}=49-2$
$x^4+\frac{1}{x^4}=47$

On squaring both sides, we get:
${(x^4+\frac{1}{x^4})}^2={47}^2$
$x^8+2\times x^4\times \frac{1}{x^4}+{\left(\frac{1}{x^4}\right)}^2=2209$
$x^8+2+\frac{1}{x^8}=2209$
$x^8+\frac{1}{x^8}=2209-2$
$x^8+\frac{1}{x^8}=2207$