Hi skoker!
It seems to me you have the right approach.
Raise both sides of your equation to the 3rd power and you get a 6th order polynomial.
It has 2 obvious roots which are the only real roots.
i am not sure how to find the points of intersection of these 2 functions algebraically.
and
i would normally do this
and solve for x. but in this case the cube is making this a nasty approach.
one try got me to here
but this seems like i have wondered into the weeds.
i guess it is the only way to raise the cube.
@MarkFL2
this substitution you did, what is it called?
if you use the normal algebra and raise both sides you get
then you start the factor.
but what you did there looks like the U-substitution from calculus for finding the anti-derivative of a composite function.
where in the algebra is this type of substitution explained? maybe for parametric equations?
Read here:
Rational Zeros of Polynomials
It is called "substitution".
Just what you already said.
If you have an equation with a difficult sub expression, you give the sub expression a name, say "u", and replace it.
Then you solve the equation, and when you're done, you back substitute "u".
Actually, substitution in integrals is a more advanced version of regular substitution.
thanks for the info...
@ILikeSerena
i will go back and look at the substitution a bit. it seems in the algebra they only give some easy examples.
like substitute for
for factoring. but not how to use with roots and more complicated expressions.
until they show how to do the u-substitution.