Question

If${\left(\frac{m}{n}\right)}^{\frac{3}{8}}+{\left(\frac{n}{m}\right)}^{\frac{3}{8}}=-18$ then find value of ${\left(\frac{m}{n}\right)}^{\frac{1}{8}}+{\left(\frac{n}{m}\right)}^{\frac{1}{8}}$

• A. $9$
• B. $-9$
• C. $3$
• D. $-3$

Answer 1

The correct option is D $-3$

Consider the given equation,

${\left(\frac{m}{n}\right)}^{\frac{3}{8}}+{\left(\frac{n}{m}\right)}^{\frac{3}{8}}=-18$ . …………(1)

We have to find, ${\left(\frac{m}{n}\right)}^{\frac{1}{8}}+{\left(\frac{n}{m}\right)}^{\frac{1}{8}}=$ (2)

Let, ${\left(\frac{m}{n}\right)}^{\frac{1}{8}}=A$, then ${\left(\frac{n}{m}\right)}^{\frac{1}{8}}=\frac{1}{A}$ (3).

Now,

From equation (1)and (3),

$A^3+\frac{1}{A^3}=-18$ ………..(4)

$\because {(A+\frac{1}{A})}^3=A^3+\frac{1}{A^3}+3A.\frac{1}{A}(A+\frac{1}{A})=A^3+\frac{1}{A^3}+3(A+\frac{1}{A})$

From equation (4), put the value of, $A^3+\frac{1}{A^3}=-18$.

$\therefore {(A+\frac{1}{A})}^3=-18+3(A+\frac{1}{A})$

Assume that ,

$A+\frac{1}{A}=B$ ……….(5)

So,

$B^3=-18+3B$

$B^3-3B+18=0$ ............(6)

Put,

$B=-3$

${(-3)}^3-3(-3)+18=0$

$-27+9+18=0$

$0=0$

Its means, $B=-3\Rightarrow B+3=0$Is the solution of equation (6).

Now, $B^3-3B+18=0$, is divided by $B+3$, we get $B^2+3B-9=0$ I.e.

$B^3-3B+18$ $=(B^2+3B-9)(B+3)$ =$0$

$B^2+3B-9\ne 0$

So,

$B+3=0$

$B=-3$

From equation (5),

$A+\frac{1}{A}=B=-3$

$A+\frac{1}{A}=-3$

From equation (3), put the value of .$A={\left(\frac{m}{n}\right)}^{\frac{1}{8}},\frac{1}{A}={\left(\frac{n}{m}\right)}^{\frac{1}{8}}=3$, we get

$A+\frac{1}{A}={\left(\frac{m}{n}\right)}^{\frac{1}{8}}+{\left(\frac{n}{m}\right)}^{\frac{1}{8}}=-3$

Hence, this is the answer.