Question

If x square + 1 upon x square=47 then find x cube + 1 upon x cube

Answer 1

Answer:

Answer is given below

Step-by-step explanation:

$$\begin{lgathered}{x}^{2} + \frac{1}{ {x}^{2} } = 47 \\ {(x + \frac{1}{x} )}^{2} - 2 = 47 \\ x + \frac{1}{x} = \sqrt{49} = 7 \: \: or \: \: - 7\end{lgathered}$$

$$\begin{lgathered}{x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x} )}^{3} - 3(x + \frac{1}{x} ) \\ = {7}^{3} - 3 \times 7 \: \: or \: \: {( - 7)}^{3} - 3 \times - 7 \\ = 322 \: \: or \: \: - 322\end{lgathered}$