Math-advanced, i think.?

Question

here is my problem, and I have tried to figure it out but i cant. The furthest i got was trying the sum of squares which didn't work out because I had a cubed left and the answer didn't fit into what they asked. So,

The transformation x=z - b/3a converts the cubic equation ax^3 + bx^2 + cx + d=0 into one of the form z^3 + 3Hz + G=0. Find H and G in terms of a, b, c, d (Someone help this is driving me crazy!!!)

Answer 1

Most of this is tedious algebra--it's really easy to accidentally drop a sign or term here or there. But if you stick it out to the end it indeed simplifies :)

Plugging in x = (z - b/3a) into the cubic equation, we get

a(z - b/3a)^3 + b(z - b/3a)^2 + c(z - b/3a) + d = 0

I'm lazy and don't want to do the cubing and squaring (but you could do it that way). Instead of explicitly cubing and squaring the first two terms, I find it easier to factor out a square term to see if terms would combine:

(z - b/3a)^2 * (a * (z - b/3a) + b) + c(z - b/3a) + d = 0
(z - b/3a)^2 * (az - b/3 + b) + c(z - b/3a) + d = 0
(z - b/3a)^2 * (az + 2b/3) + c(z - b/3a) + d = 0

And they do! So then I factor out another (z - b/3a) term, and multiply to see if more stuff combines:

(z - b/3a) * ( (z - b/3a)(az + 2b/3) + c) + d = 0
(z - b/3a) * ( az^2 + 2bz/3 - bz/3 - 2b^2/9a + c) + d = 0
(z - b/3a) * ( az^2 + bz/3 - 2b^2/9a + c) + d = 0

Now I actually multiply everything out:

(z - b/3a) * ( az^2 + bz/3) + (z - b/3a)(-2b^2/9a + c) + d = 0
az^3 + bz^2/3 - bz^2/3 - b^2z/9a + (z - b/3a)(-2b^2/9a + c) + d = 0

And indeed the z^2 term goes away. Simplifying more:

az^3 - (b^2/9a)z + (-2b^2z/9a + cz + 2b^3/27a^2 - bc/3a) + d = 0
az^3 -b^2z/3a + cz + 2b^3/27a^2 - bc/3a + d = 0
az^3 + z( -b^2/3a + c) + (2b^3/27a^2 - bc/3a + d) = 0

Since we want to get things in the form z^3 + 3Hz + G = 0, divide by a:

z^3 + z(-b^2/3a^2 + c/a) + (2b^3/27a^3 - bc/3a^2 + d/a) = 0
So 3H = (-b^2/3a^2 + c/a)
H = -b^2/9a^2 + c/3a
H = -(b/3a)^2 + c/3a
G = 2*(b/3a)^3 - c/a * (b/3a) + d/a
G = 2(b/3a)^3 - bc/3a^2 + d/a

Answer 2

x^3 = z^3 - 3bz^2/3a + 3b^2z/9a^2 + b^3/27a^2
x^2 = z^2 - 2bz/3a + b^2z/9a^2
ax^3 =az^3-bz^2 +b^2z/3a +b^3/27a
bx^2 = bz^2 -2b^2z/3a +b^3z/9a^2
cx = cz -bc/3a
d = d
When we add thes up we see that the bz^2 term is gone and we have the following:
az^3-b^2z/3a+b^3z/9a^2+cz +b^3/27a -bc/3a +d
So 3H = -b^2/3a +b^3/9a^2 +c
and G = b^3/27a-bc/3a +d