Use linear approximation, i.e. the tangent line, to approximate 1.6^3 as follows:

Let f(x) = x ^3. The equation of the tangent line to f(x) at x = 2 can be written as y=12x-16
Using this, we find our approximation for 1.6 ^3 is ???????

To use linear approximation (the tangent line) to approximate 1.6^3, we need to substitute the value x = 1.6 into the equation of the tangent line.

The equation of the tangent line to f(x) = x^3 at x = 2 is given as y = 12x - 16.

So, let's substitute x = 1.6 into the equation:
y = 12(1.6) - 16
y = 19.2 - 16
y = 3.2

Therefore, using linear approximation, 1.6^3 is approximated to be 3.2.