Question

x^4+1/x^4=23 , x^3-1/x^3=?

Answer 1

$\displaystyle \Longrightarrow\ x^4+\frac{1}{x^4}=23 \\ \\ \\ \\ \Longrightarrow\ x^4+\frac{1}{x^4}+2=23+2 \\ \\ \\ \\ \Longrightarrow\ (x^2)^2+\left(\frac{1}{x^2}\right)^2+\left(2 \times x^2 \times \frac{1}{x^2}\right)=25 \\ \\ \\ \\ \Longrightarrow\ \left(x^2+\frac{1}{x^2}\right)^2=25 \\ \\ \\ \\ \Longrightarrow\ x^2+\frac{1}{x^2}=\ \pm 5$

$\displaystyle \textsf{1. Taking \ x^2+\frac{1}{x^2}=5 ,} \\ \\ \\ \\ \Longrightarrow\ x^2+\frac{1}{x^2}-2=5-2 \\ \\ \\ \\ \Longrightarrow\ (x)^2+\left(\frac{1}{x}\right)^2-\left(2 \times x \times \frac{1}{x}\right)=3 \\ \\ \\ \\ \Longrightarrow\ \left(x-\frac{1}{x}\right)^2=3 \\ \\ \\ \\ \Longrightarrow\ x-\frac{1}{x}=\pm \sqrt{3}$

$\displaystyle \textsf{(i) Taking \ x-\frac{1}{x}=\sqrt{3} ,} \\ \\ \\ \\ \Longrightarrow\ \left(x-\frac{1}{x}\right)^3=(\sqrt{3})^3 \\ \\ \\ \\ \Longrightarrow\ x^3-3x+\frac{3}{x}-\frac{1}{x^3}=3\sqrt{3} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)=3\sqrt{3} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}-3\sqrt{3}=3\sqrt{3} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}=3\sqrt{3}+3\sqrt{3} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}=\Large \ \text{ \bold{6\sqrt{3}} }$

$\displaystyle \textsf{(ii) Taking \ x-\frac{1}{x}=-\sqrt{3} ,} \\ \\ \\ \\ \Longrightarrow\ \left(x-\frac{1}{x}\right)^3=(-\sqrt{3})^3 \\ \\ \\ \\ \Longrightarrow\ x^3-3x+\frac{3}{x}-\frac{1}{x^3}=-3\sqrt{3} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)=-3\sqrt{3} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}+3\sqrt{3}=-3\sqrt{3}$

$\displaystyle \Longrightarrow\ x^3-\frac{1}{x^3}=-3\sqrt{3}-3\sqrt{3} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}=\Large \ \text{ \bold{-6\sqrt{3}} }$

$\displaystyle \textsf{2. Taking \ x^2+\frac{1}{x^2}=-5 ,} \\ \\ \\ \\ \Longrightarrow\ x^2+\frac{1}{x^2}-2=-5-2 \\ \\ \\ \\ \Longrightarrow\ (x)^2+\left(\frac{1}{x}\right)^2-\left(2 \times x \times \frac{1}{x}\right)=-7 \\ \\ \\ \\ \Longrightarrow\ \left(x-\frac{1}{x}\right)^2=-7 \\ \\ \\ \\ \Longrightarrow\ x-\frac{1}{x}=\pm \sqrt{-7} \\ \\ \\ \\ \Longrightarrow\ x-\frac{1}{x}=\pm \iota \sqrt{7}$

$\displaystyle \textsf{(i) Taking \ x-\frac{1}{x}=\iota\sqrt{7} ,} \\ \\ \\ \\ \Longrightarrow\ \left(x-\frac{1}{x}\right)^3=(\iota\sqrt{7})^3 \\ \\ \\ \\ \Longrightarrow\ x^3-3x+\frac{3}{x}-\frac{1}{x^3}=-7\iota\sqrt{7} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)=-7\iota\sqrt{7} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}-3\iota\sqrt{7}=-7\iota\sqrt{7}$

$\displaystyle \Longrightarrow\ x^3-\frac{1}{x^3}=-7\iota\sqrt{7}+3\iota\sqrt{7} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}=\Large \ \text{ \bold{-4\iota\sqrt{7}} }$

$\displaystyle \textsf{(ii) Taking \ x-\frac{1}{x}=-\iota\sqrt{7} ,} \\ \\ \\ \\ \Longrightarrow\ \left(x-\frac{1}{x}\right)^3=(-\iota\sqrt{7})^3 \\ \\ \\ \\ \Longrightarrow\ x^3-3x+\frac{3}{x}-\frac{1}{x^3}=7\iota\sqrt{7} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)=7\iota\sqrt{7} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}+3\iota\sqrt{7}=7\iota\sqrt{7}$

$\displaystyle \Longrightarrow\ x^3-\frac{1}{x^3}=7\iota\sqrt{7}-3\iota\sqrt{7} \\ \\ \\ \\ \Longrightarrow\ x^3-\frac{1}{x^3}=\Large \ \text{ \bold{4\iota\sqrt{7}} }$

$\textsf{Thus the 4 possible answers are, } \\ \\ \\ \Large \text{ \bold{1.\ \ \ \ \ \ 6\sqrt{3}} } \\ \\ \text{ \bold{2.\ \ -6\sqrt{3}} } \\ \\ \text{ \bold{3.\ \ -4\iota\sqrt{7}} } \\ \\ \text{ \bold{4.\ \ \ \ \ \ 4\iota\sqrt{7}} }$