Question

if x^4+1/x^4=47
#then x^3+1/x^3=,?​

Answer 1

Answer:

${x}^{4} + \frac{1}{ {x}^{4} } = 47$

$( {x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} - 2 {x}^{2} \times \frac{1}{ {x}^{2} } = 47$

$( {x}^{2} + \frac{1}{ {x}^{2} } ) = \sqrt{47 + 2} = \sqrt{49} = 7$

$(x + \frac{1}{x} ) {}^{2} - 2x \times \frac{1}{x} = 7$

$x + \frac{1}{x} = \sqrt{7 + 2} = \sqrt{9} = 3$

now,

${x}^{3} + \frac{1}{ {x}^{3} } = (x + \frac{1}{x} ) {}^{3} - 3x \frac{1}{x} (x + \frac{1}{x} ) = {3}^{3} - 3 \times 3 = 27 - 9 = 18 \: \: ans.$