Attempt to solve f(y) = -5y^3+6y^2-4 = 0f(y)=−5y3+6y2−4=0
First divide through by -y^3−y3 to get:
5-6/y+4/y^3 = 05−6y+4y3=0
Let x = 1/yx=1y
Then 4x^3-6x+5 = 04x3−6x+5=0
Now let x = u + vx=u+v
0 = 4(u+v)^3 - 6(u+v) + 50=4(u+v)3−6(u+v)+5
=4u^3+4v^3+(12uv-6)(u+v)+5=4u3+4v3+(12uv−6)(u+v)+5
=4u^3+4v^3+6(2uv-1)(u+v)+5=4u3+4v3+6(2uv−1)(u+v)+5
Let v = 1/(2u)v=12u
=4u^3+1/(2u^3)+5=4u3+12u3+5
Multiply through by 2u^32u3 to get:
8(u^3)^2+10(u^3)+1 = 08(u3)2+10(u3)+1=0
u^3 = (-10+-sqrt(100-32))/16u3=−10±√100−3216
=(-10+-sqrt(68))/16=−10±√6816
=(-5+-sqrt(17))/8=−5±√178
Write:
u_1 = root(3)((-5+sqrt(17))/8)u1=3√−5+√178
v_1 = root(3)((-5-sqrt(17))/8)v1=3√−5−√178
Then real root of 4x^3-6x+5 = 04x3−6x+5=0 is
x = u_1 + v_1x=u1+v1
The other two (complex) roots are:
x = omega u_1 + omega^2 v_1x=ωu1+ω2v1
x = omega^2 u_1 + omega v_1x=ω2u1+ωv1
where omega = -1/2+sqrt(3)/2iω=−12+√32i
y = 1/xy=1x
So the real root of f(y) = 0f(y)=0 is y_1 = 1/(u_1+v_1)y1=1u1+v1
and the complex roots are:
y_2 = 1/(omega u_1 + omega^2 v_1)y2=1ωu1+ω2v1
y_3 = 1/(omega^2 u_1 + omega v_1)y3=1ω2u1+ωv1
f(y) = -5(y - y_1)(y - y_2)(y - y_3)f(y)=−5(y−y1)(y−y2)(y−y3)
