Question

154) Find all possible values of x^3 + 1/x^3 given that x^2+ 1/X^2 = 7.

Answer 1

Solving for x in the equation, we have:

`\begin{gathered} x^2+(1)/(x^2)=7 \\ x^4+1=7^{}x^2(^{}\text{ Multiplying by x}^2\text{ on both sides of the equation and distributing)} \\ x^4-7^{}x^2+1=0(\text{ Subtracting 7x}^2\text{ from both sides of the equation)} \end{gathered}`

`\begin{gathered} y^2-7^{}y+1=0\text{ (Replacing y=x}^2\text{ in the equation)} \\ \text{ Using the quadratic equation with a=1, b=-7 and c=1, we have:} \\ \frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \frac{-(-7)\pm\sqrt[]{(-7)^2-4(1)(1)}}{2(1)}\text{ (Replacing the values)} \end{gathered}`

`\begin{gathered} \frac{7\pm\sqrt[]{49-4}}{2}\text{ (Using the sign rules, raising -7 to the power of 2 and multiplying)} \\ y1=\frac{7+\sqrt[]{45}}{2}=6.85\text{ (Simplifying to find the first solution)} \\ y2=\frac{7-\sqrt[]{45}}{2}=0.146\text{ (Simplifying to find the second solution)} \end{gathered}`

`\begin{gathered} \text{ Given that y=x}^2,\text{ we have to find the values of x. Doing so, we have:} \\ x1=\sqrt[]{6.85}=2.62 \\ x2=-\sqrt[]{6.85}=-2.62 \\ x3=\sqrt[]{0.146}=0.382 \\ x4=-\sqrt[]{0.146}\text{ =}-0.382 \end{gathered}`

Replacing the previous values in the expression x^3 + 1/(x^3), we have:

`\begin{gathered} \text{First value: (2.62)}^{}^3+(1)/((2.62)^3)=18.04 \\ \text{ Second value: (-2.62)}^3+(1)/((-2.62)^3)=-18.04 \\ \text{ Third value: (0.382)}^3+(1)/((0.382)^3)=17.995 \\ \text{ Fourth value: (-0.382)}^3+(1)/((-0.382)^3)=-17.995 \end{gathered}`