Cubic functions- solve for x?

Question

I am having trouble with finding the roots of cubic functions by solving for x on problems with missing parts such as: x^3 - x, and: x^3 - 1,

Can anyone help me?

Answer 1

So you want to find the roots of x^3 - x and x^3 - 1. These are actually the simplest cubics to solve (which is fortunate because solving a full-fledged cubic such as x^3 + 5x^2 + 2x + 3 is no easy matter).

If x^3 - x = 0, then all you have to do is factor it.

x^3 - x = 0
x(x^2 - 1) = 0, therefore we get factors
x(x - 1)(x + 1) = 0, meaning
x = {0, 1, -1}

If x^3 - 1 = 0, there are two ways to do this; one that avoids the complex roots, and one that doesn't. I'll show you the one that avoids the complex roots first.

x^3 - 1 = 0 means
x^3 = 1, and taking the cube root of both sides, we get
x = 1.

This method eliminates getting the complex roots though and isn't generally recommended when taking university-level math. The proper way to do this is this by factoring it as a difference of cubes. Recall that if (a^3 - b^3), then this factors into
(a - b) (a^2 + ab + b^2). We do the same thing

x^3 - 1 = 0
(x - 1) (x^2 + x + 1) = 0, which becomes
x = 1 and we have to use the quadratic formula on the second.

x^2 + x + 1 = 0
x = [-1 +/- sqrt (1 - 4)]/2
x = [-1 +/- sqrt (-3)]/2
x = -1/2 +/- (1/2) sqrt(-3)
x = -1/2 +/- (1/2) sqrt(3)i
{note that i = sqrt(-1)

Answer 2

1) x^3-x -----> it can be rewritten as x(x^2-1) since x is common for both. now equate it to zero....it becomes....x=0 and x^2-1=0, which inturn becomes x^2=1.....x=sqrt(1). so the answer is

0,+1 and -1

2) x^3 - 1 ----> first equate it to zero...x^3 - 1=0. it becomes
x^3 = 1.....therefore the answer for this Q is 1,1 and 1..

note:this is because, it is a cube root and thus no negative roots.

Answer 3

I think i know the ans fow yhe 2nd one i.e x^3-1.u can cube 1 as it returns the same value 1^3=1.therefore x^3-1^3.And then u simplify it.

Answer 4

that's considered one of those question i like. My Dad is a scientist specialising in quantum physics. Er, the conventional variety of cubic function? Er, This h(x)= stuff. Are you protecting down the keys too long? Like PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP? i think of it truly is factoring. could desire to be quadratic yet y'understand, one formulation is like yet another in some respects- it truly is totally formulaic. besides, desire this enables.

Answer 5

x^3-x=x(x+1)(x-1)
the roots are 0,+/-1
x^3-1=(x-1)(x^2+x+1)
the rootsare x=1 0r[-1+/-rt(1-4)]/2
=1 or (-1/2)+/-irt3/2