MHB How do I simplify this calculus problem?

[tmt1]

Hi I'm working on a problem for calculus

and I got this

2x + 3 (y-x)^2 * (y'-1) = 0 .

And then, in the answer part it says I have to get to here, which I'm not sure how to do.

2x + 3 (y-x)^2 * y'- 3 (y-x)^2 = 0 ,

Thanks,

Tim

 

[Prove It]

Hi I'm working on a problem for calculus

and I got this

2x + 3 (y-x)^2 * (y'-1) = 0 .

And then, in the answer part it says I have to get to here, which I'm not sure how to do.

2x + 3 (y-x)^2 * y'- 3 (y-x)^2 = 0 ,

Thanks,

Tim

It's \displaystyle \begin{align*} 3a\left( b - c \right) = 3a\,b - 3a\,c \end{align*}, where \displaystyle \begin{align*} a = \left( y - x \right) ^2 , b = y', c = 1 \end{align*}.