Question

Prove that a cubic equation x 3 + ax 2 + bx+ c = 0 has 3 roots by finding the roots.

Answer 1

That's a pretty tall order for Brainly homework.  Let's start with the depressed cubic, which is simpler.

Solve

$y^3 + 3py = 2q$

We'll put coefficients on the coefficients to avoid fractions down the road.

The key idea is called a split, which let's us turn the cubic equation in to a quadratic.  We split unknown y into two pieces:

$y = s + t$

Substituting,

$(s+t)^3 + 3p(s+t) = 2q$

Expanding it out,

$s^3+3 s^2 t + 3 s t^2 + t^3 + 3p(s+t) = 2q$

$s^3+t^3 + 3 s t(s+t) + 3p(s+t) = 2q$

$s^3+t^3 + 3( s t + p)(s+t) = 2q$

There a few moves we could make from here. The easiest is probably to try to solve the simultaneous equations:

$s^3+t^3=2q, \qquad st+p=0$

which would give us a solution to the cubic.

$p=-st$

$t = -\dfrac p s$

Substituting,

$s^3 - \dfrac{p^3}{s^3} = 2q$

$(s^3)^2 - 2 q s^3 - p^3 = 0$

By the quadratic formula (note the shortcut from the even linear term):

$s^3 = q \pm \sqrt{p^3 + q^2}$

By the symmetry of the problem (we can interchange s and t without changing anything) when s is one solution t is the other:

$s^3 = q + \sqrt{p^3+q^2}$

$t^3 = q - \sqrt{p^3+q^2}$

We've arrived at the solution for the depressed cubic:

$y = s+t = \sqrt[3]{q + \sqrt{p^3+q^2}} + \sqrt[3]{ q - \sqrt{p^3+q^2} }$

This is all three roots of the equation, given by the three cube roots (at least two complex), say for the left radical.  The two cubes aren't really independent, we need their product to be $-p=st$.

That's the three roots of the depressed cubic; let's solve the general cubic by reducing it to the depressed cubic.

$x^3 + ax^2 + bx + c=0$

We want to eliminate the squared term.  If substitute x = y + k we'll get a 3ky² from the cubic term and ay² from the squared term; we want these to cancel so 3k=-a.

Substitute x = y - a/3

$(y - a/3)^3 + a(y - a/3)^2 + b(y - a/3) + c = 0$

$y^3 - ay^2 + a^2/3 y - a^3/27 + ay^2-2a^2y/3 + a^3/9 + by - ab/3 + c =0$

$y^3 + (b - a^2/3) y = -(2a^3+9ab) /27$

Comparing that to

$y^3 + 3py = 2q$

we have $p = (3b - a^2) /9, q =-(a^3+9ab)/54$

which we can substitute in to the depressed cubic solution and subtract a/3  to get the three roots.  I won't write that out; it's a little ugly.