Question

Solve : $8x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2n$}\right.}-8x^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2n$}\right.}=63.$

Answer 1

The given expression is $8x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2n$}\right.}-8x^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2n$}\right.}=63.$

Step 1: put $x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2n$}\right.}=t$

$8x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2n$}\right.}-8x^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2n$}\right.}=63$

$\Rightarrow 8t-\frac{8}{t}=63$

$$\begin{gathered} \Rightarrow 8t^2-63t-8=0 \\ \Rightarrow 8t^2-64t+t-8=0 \\ \Rightarrow 8t(t-8)+1(t-8)=0 \\ \Rightarrow (8t+1)(t-8)=0 \\ \Rightarrow t=\frac{-1}{8},8 \end{gathered}$$

We have taken $x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2n$}\right.}=t$

$\Rightarrow x=t^{\raisebox{1ex}{$2n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Step 2: Finding the value of $x$ when $t=-\frac{1}{8}$

For $t=-\frac{1}{8}$

$$\begin{gathered} \Rightarrow x={\left(\frac{-1}{8}\right)}^{\raisebox{1ex}{$2n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \Rightarrow x={\left(\frac{-1}{2}\right)}^{3\times \frac{2n}{3}} \\ \Rightarrow x={\left(\frac{-1}{2}\right)}^{\cancel{3}\times \frac{2n}{\cancel{3}}} \\ \Rightarrow x={\left(\frac{-1}{2}\right)}^{2n} \\ \Rightarrow x=\left(\frac{1}{2^{2n}}\right) \end{gathered}$$

Step 3: Finding the value of $x$ when $t=8$

For $t=8$

$$\begin{gathered} \Rightarrow x={\left(8\right)}^{\raisebox{1ex}{$2n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \Rightarrow x={\left(2\right)}^{3\times \frac{2n}{3}} \\ \Rightarrow x={\left(2\right)}^{\cancel{3}\times \frac{2n}{\cancel{3}}} \\ \Rightarrow x={\left(2\right)}^{2n} \end{gathered}$$

Hence, the value of $x$ are $\frac{1}{2^{2n}},2^{2n}$ respectively.