The values of ‘b’ for which the line y=12x+b is a tangent to the curve $y=x^3$ are

The given curve is $y=x^3$

Differentiate both sides with respect to $x$

$\frac{dy}{dx}=3x^2$

Given that $y=12x+b$ is tangent to the above curve 

$\begin{array}{rcl}3x^2& =& 12\\ x^2& =& 4\end{array}$

It implies that $x=\pm 2$, hence, $y=\pm 8$

$\pm 8=12\left(\pm 2\right)+b\Rightarrow b=\boxed{\mathbf{\pm }\mathbf{16}}$