Question

Find $r$ in $a[\frac{1}{r^3}+\frac{1}{r}+r+r^3]=85$ where $a=8$.

Answer 1

We have,

$a[\frac{1}{r^3}+\frac{1}{r}+r+r^3]=85$

Since,

$a=8$

Therefore,

$8[\frac{1}{r^3}+\frac{1}{r}+r+r^3]=85$

$8[r^3+\frac{1}{r^3}+r+\frac{1}{r}]=85$

$8(r^3+\frac{1}{r^3})+8(r+\frac{1}{r})=85$

$8[{(r+\frac{1}{r})}^3-3(r+\frac{1}{r})]+8(r+\frac{1}{r})=85$

$8{(r+\frac{1}{r})}^3-24(r+\frac{1}{r})+8(r+\frac{1}{r})=85$

$8{(r+\frac{1}{r})}^3-16(r+\frac{1}{r})-85=0$ $..........\left(1\right)$

Let

$x=(r+\frac{1}{r})$

So,

$8x^3-16x-85=0$

Now, factorize this equation,

$(2x-5)(4x^2+10x+17)=0$

Since,

$4x^2+10x+17=0$ has imaginary roots.

So,

$2x-5=0$

$x=\frac{5}{2}$

Therefore,

$(r+\frac{1}{r})=\frac{5}{2}$

$2r^2-5r+2=0$

So,

$r=2$ or $\frac{1}{2}$

Hence, this is the answer.