Submitted by Some.Username on Fri, 08/03/2018 - 15:47
Forums:
I'm fine for the first two parts of the question but for the last part, I'm just wondering how it was found in the solutions that y=2x was a suitable substitution to make so that the equation would work. Is there a method or should you be able to know by just looking at it or by trial and error? Thanks.
Substitution
It's a bit trial and error-ish.
One of the approaches you can make it to start by noting that the cubics have the same structure (no squared terms etc). This means that a substitution of the form $y=kx$ (rather than $y=kx+c$) might be suitable.
If you substitute $y=kx$ into the second equation you get:
\[
y^3-3y-\sqrt{2}=0\\
k^3x^3 - 3kx - \sqrt{2} =0\
\]
Then we want to divide by $k$ so that the coefficient of $x$ is correct which gives:
\[
k^2x^3-3x - \tfrac 1 k \sqrt{2} = 0
\]
Finally, comparing this to the first equation in $x$ suggests that $k=2$ is a sensible thing to try.