Question

if x²-4x+1=0 then find the x³+1/x³​

Answer 1

$\textbf {\underline {\underline {ANSWER }}}$

${x}^{2} - 4x + 1 = 0 \\ \\ \\ {b}^{2} - 4ac = {( - 4)}^{2} - 4 \\ \\ \\ = 16 - 4 \\ \\ \\ = 12 \\ \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \\ = \frac{ - ( - 4) + - \sqrt{12} }{2} \\ \\ \\ = \frac{4 + - 2 \sqrt{3} }{2} \\ \\ \\ x = 2 + \sqrt{3} \: \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: 2 - \sqrt{3} \\ \\ \\ x + \frac{1}{x} = 2 + \sqrt{3} + \frac{1}{2 + \sqrt{3} } \\ \\ \\ = 2 + \sqrt{3} + \frac{ 2 - \sqrt{3} }{(2 - \sqrt{3)}(2 + \sqrt{3} )} \\ \\ \\ = 2 + \sqrt{3} + \frac{2 - \sqrt{3} }{ - 1} \\ \\ \\ = 2 + \sqrt{3} - 2 + \sqrt{3} \\ \\ \\ = 2 \sqrt{3} \\ \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x} )}^{3} - 3(x + \frac{1}{x} ) \\ \\ \\ = {(2 \sqrt{3} )}^{3} - 3(2 \sqrt{3)} \\ \\ \\ = 72 \sqrt{3} - 6 \sqrt{3} \\ \\ \\ = 66 \sqrt{3} \\ \\ \\ x = 2 - \sqrt{3} \\ \\ \\ x + \frac{1}{x} = 2 - \sqrt{3} + \frac{1}{2 - \sqrt{3} } \\ \\ \\ = 2 - \sqrt{3} + \frac{2 + \sqrt{3} }{(2 + \sqrt{3})(2 - \sqrt{3} ) } \\ \\ \\ = 2 - \sqrt{3} - 2 - \sqrt{3} \\ \\ \\ = 0$

Thanks

Answer 2

Solution:-

Given Quadratic Polynomial Equation:-

=) x² - 4x + 1 = 0

=) x² - 4x = -1

=) x ( x - 4) = -1

=) x -4 = -1/x

=) x + 1/x = 4 ______________(1)

Now,

Taking Cube on both the sides. we get,

=) ( x + 1/x )³ = 4³

=) x³ + (1/x)³ + 3.x.1/x. ( x + 1/x) = 64

=) x³ + 1/x³ + 3 ( x + 1/x ) = 64

=) x³ + 1/x³ + 3 × 4 = 64 [ From eq(1). ]

=) x³ + 1/x³ = 64 - 12

=) x³ + 1/x³ = 52

Hence,

x³ + 1/x³ = 52

Identity Used:-

• ( a + b)³ = a³ + b³ + 3ab ( a + b).