Question

x²-3x-1=0
find x³+1/x³ fast

Answer 1

You did mistake
here question may be x² -3x + 1 = 0
Then, find x³ + 1/x³ = ?

x² - 3x + 1 = 0
dividing by x both sides,
x²/x - 3x/x + 1/x = 0/x
x - 3 + 1/x = 0
x + 1/x = 3
Now,
x³ + 1/x³ = (x + 1/x)³ - 3x.1/x(x + 1/x)
= (3)³ - 3(3) = 27 - 9 = 18

Hence, 18 is answer .

Answer 2

Heya User,

--> x² - 3x - 1 = 0
=> x² - 2[x][3/2] + [3/2]² - [3/2]² - 1 = 0
=> [ x - 3/2 ]² = 1 + 9/4
=> [ x - 3/2  ]² = 13/4
=> [ x - 3/2 ] = + [ √13 / 2 ] ; - [ √13 / 2 ]
=> x = [ 3 ± √13 ] /2

=> 1/x = 2 / [ 3 ± √13 ]
=> $\frac{1}{x} = \frac{2[3^-_+ \sqrt{13} ]}{9-13} = \frac{2[3^-_+ \sqrt{13}]}{-4}= \frac{[\sqrt{13}^-_+ 3]}{2}$
=> x + 1/x = $\frac{[3^+_- \sqrt{13}] }{2} + \frac{[\sqrt{13}^-_+ 3]}{2} = \sqrt{13}\:or\:-\sqrt{13}$

[tex][x + \frac{1}{x} ]^3 = x^3 + \frac{1}{x^3} + 3[x + \frac{1}{x} ]\\ =\ \textgreater \ ^+_-13 \sqrt{13} = x^3 + \frac{1}{x^3} + 3^+_- \sqrt{13}\\ =\ \textgreater \ x^3 + \frac{1}{x^3} = ^+_-10 \sqrt{13}[/tex]

Hence, x + 1/x = ±10√13