The equation x^3 - 3x + 8 = 0 is a cubic equation of the form ax^3 + bx^2 + cx + d = 0 where a = 1, b = 0, c = -3 and d = 8

The number of real roots of the equation is determined by evaluating the value of 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2

18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2 = 0 + 0 + 108 - 1728 = -1620 < 0

The equation has only one real root.

The root can be determined in terms of a, b, c and d. As b = 0, in the following expression all terms with b have been eliminated.

X = `-1/(3a)*root(3)((1/2)*(27a^2d + sqrt((27a^2d)^2 - 4*(-3ac)^3)))`

`- 1/(3a)*root(3)((1/2)*(27a^2d - sqrt((27a^2d)^2 - 4*(-3ac)^3)))`

=> `(-1/3)*root(3)((1/2)*(27*8 + sqrt((27*8)^2 - 4*9^3)))`

`(-1/3)*root(3)((1/2)*(27*8 - sqrt((27*8)^2 - 4*9^3)))`

=> `(-1/3)*root(3)((1/2)*(216 + sqrt(216^2 - 2916)))`

`(-1/3)*root(3)((1/2)*(216 - sqrt(216^2 - 2916)))`

=> `(-1/3)*root(3)((1/2)*(216 + sqrt(216^2 - 2916)))`

` - (-1/3)*root(3)((1/2)*(216 - sqrt(216^2 - 2916)))`

=> `-root(3)(4 + sqrt 15) - root(3)(4 - sqrt 15)`

**The real solution of the equation x^3 - 3x + 8 = 0 is `-root(3)(4 + sqrt 15) - root(3)(4 - sqrt 15)`**

To find the x-intercept of y=x^3-3x+8, yes you need to solve

Y=x^3-3x+8=0 -----(1)

It can be solved graphically by plotting the above equation and looking at what **"x"** value the function becomes "zero"

Approximately y becomes 0 at x = - 2.5

so x intercept is (0 - (-2.5) = 2.5