#### Answer

The tangent lines to $y$ at $x=a$ and $x=b$ are parallel when $a=b$ or $a+b=2$

#### Work Step by Step

The slope of the tangent line is
$\frac{dy}{dx}=3\times\frac{1}{3}x^{3-1}-2x^{2-1}=x^{2}-2x$
The tangent lines to $y$ at $x=a$ and at $x=b$ are parallel if the slopes are the same. That is, if $\frac{dy}{dx}|_{x=a}=\frac{dy}{dx}|_{x=b}$
If $a=b$,
$\frac{dy}{dx}|_{x=a}=a^{2}-2a=b^{2}-2b=\frac{dy}{dx}|_{x=b}$
If $a+b=2$,
$\frac{dy}{dx}|_{x=a}=a^{2}-2a$
and as $b=2-a$,
$\frac{dy}{dx}|_{x=b}=(2-a)^{2}-2(2-a)$
$=4+a^{2}-4a-4+2a=a^{2}-2a$
By proving that the slopes are equal, we showed that the tangent lines to $y$ at $x=a$ and $x=b$ are parallel when $a=b$ or $a+b=2$