Question

if x square - 5 x + 1 is equal to zero then find the value of x cube + 1 by x cube​

Answer 1

x^2 - 5x + 1 = 0

x * x - 5x = 0-1 = -1

x - 4x = -1

3x = -1

x = -1/3

x^3 + 1/x^3

-1/3^3 + 1/-1/3^3

-1/9 + 1/-1/9

=> -1/9 -1/9 +1

=> -2/9 + 1

=> -2-9/9

=> -11/9

=> 1.22

Answer 2

Answer:

$x^3+\frac{1}{x^3}\approx 109$    

Step-by-step explanation:

Given : If $x^2-5x+1=0$

To find : The value of $x^3+\frac{1}{x^3}$ ?

Solution :

The quadratic equation $x^2-5x+1=0$ is solve by quadratic formula,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$            

Here, a=1 , b=-5, c=1

$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(1)}}{2(1)}$

$x=\frac{5\pm\sqrt{25-4}}{2}$

$x=\frac{5\pm\sqrt{21}}{2}$

$x=\frac{5+\sqrt{21}}{2},\frac{5-\sqrt{21}}{2}$

$x=4.791,0.209$

At x=4.791,

$x^3+\frac{1}{x^3}=(4.791)^3+\frac{1}{(4.791)^3}$

$x^3+\frac{1}{x^3}=109.97+\frac{1}{109.97}$

$x^3+\frac{1}{x^3}=109.98$

At x=0.209,

$x^3+\frac{1}{x^3}=(0.209)^3+\frac{1}{(0.209)^3}$

$x^3+\frac{1}{x^3}=0.009129+\frac{1}{0.009129}$

$x^3+\frac{1}{x^3}=109.54$