Question
If $x^2+\frac{1}{x^2}=34$, find $x^3+\frac{1}{x^3}-9$.
✓
Answer
As we know
$\left(x+\frac{1}{x}\right)^2$
$=x^2+\frac{1}{x^2}+2$
$=34+2$
$=36$
$\left(x+\frac{1}{x}\right)=6$
On cubing both sides, we get
$\left(x+\frac{1}{x}\right)^3=6^3$
$\Rightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=216$
$\Rightarrow x^3+\frac{1}{x^3}+3 \times 6=216$
$\Rightarrow x^3+\frac{1}{x^3}=198$
$\Rightarrow x^3+\frac{1}{x^3}-9$
$=198-9$
$=189$
$\left(x+\frac{1}{x}\right)^2$
$=x^2+\frac{1}{x^2}+2$
$=34+2$
$=36$
$\left(x+\frac{1}{x}\right)=6$
On cubing both sides, we get
$\left(x+\frac{1}{x}\right)^3=6^3$
$\Rightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=216$
$\Rightarrow x^3+\frac{1}{x^3}+3 \times 6=216$
$\Rightarrow x^3+\frac{1}{x^3}=198$
$\Rightarrow x^3+\frac{1}{x^3}-9$
$=198-9$
$=189$
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