Question

if
${x}^{2 } - 4x + 1 = 0$
find
${x}^{3} + \frac{1}{ {x}^{3} }$
​

Answer 1

Answer:

your question is wrong

Answer 2

answer :-52

$given - - > \\ \\ {x}^{2} - 4x + 1 = 0 \\ \\ {x}^{2} + 1 = 4x \\ \\ x(x + \frac{1}{x} ) = 4x \\ \\ so \\ \\ (x + \frac{1}{x} ) = 4 \\ \\ now \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x} )}^{3} - 3(x + \frac{1}{x} \\ \\ = {(4)}^{3} - 3(4) \\ \\ = 64 - 12 = 52$