Question

if x +1/x = root 2 find (a) x3 +1/x3 (b) x2 + 1/x2

Answer 1

Answer:

a=-√2

b=0

Step-by-step explanation:

$x + \frac{1}{x} = \sqrt[]{2} \\ cube \: both \: side \: we \: get \\ {(x + \frac{1}{x} )}^{3} = { \sqrt[]{2} }^{3 } \\ = {x}^{3} + \frac{1}{ {x}^{3} } + 3x \times \frac{1}{x} (x + \frac{1}{x} ) = 2 \sqrt{2} \\ = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times \sqrt{2} = 2 \sqrt{2} \\ = {x}^{3} + \frac{1}{ {x}^{3} } = 2 \sqrt{2 } - 3 \sqrt{2} = - \sqrt{2} \\ \\ now \: square \: bth \: side \: the \: first \: equatin \: \\ {(x + \frac{1}{x} )}^{2} = {( \sqrt{2}) }^{2} \\ = {x}^{2} + \frac{1}{ {x}^{2} } + 2x \times \frac{1}{x} = 2 \\ = {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 2 \\ = {x}^{2} + \frac{1}{ {x}^{2} } = 2 - 2 = 0$