First, determine the slope of the tangent line. Take note that the slope of a line tangent to the curve is equal to the derivative of the function at the point of tangency. So taking the derivative of the function, f'(x) will be:

`f'(x) = d/dx (x^3)`

`f'(x) = 3x^2`

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`f'(x) = 3x^2`

The given point of tangency is (1,1). Plugging in x=1 to the derivative yields:

`f'(1) = 3(1)^2`

`f'(1) = 3`

Hence, the slope of the line tangent to the function at (1,1) is m = 3.

Then, apply the point slope-form to get the equation of the tangent line.

`y - y_1 =m(x - x_1)`

`y - 1 = 3(x - 1)`

Isolating the y, it becomes:

`y - 1 =3x - 3`

`y = 3x - 2`

Thus, the equation of the tangent line is `y = 3x - 2` .

Then, determine the intersection points of `y = x^3` and `y = 3x - 2`. To do so, set the two y's equal to each other.

`y = y`

`x^3 = 3x - 2`

Take note that to solve polynomial equation, one side should be zero.

`x^3 - 3x + 2 = 0`

Then factor the left side using grouping method.

`x^3 - x - 2x + 2=0`

`(x^3 - x) + (-2x + 2) = 0`

`x(x^2 - 1) - 2(x - 1) = 0`

`x(x-1)(x+1) - 2(x - 1) = 0`

`(x - 1)[x(x + 1) - 2] = 0`

`(x - 1)(x^2+x-2)=0`

`(x - 1)(x-1)(x + 2)=0`

`(x-1)^2(x + 2) = 0`

Set each factor equal to zero.

- `(x-1)^2 = 0`

`x - 1=0`

`x=1`

- `x + 2=0`

`x=-2`

Then, plug-in the x values to either y=x^3 or y = 3x - 2, to get the y coordinates of the intersection.

`x=1`

`y = 1^3=1`

`x=-2`

`y=(-2)^3 = -8`

So, the two equations intersect at (1,1) and (-2,-8). Hence, bounded region of y = x^3 and y = 3x - 2 is:

To determine the area of the bounded region, draw a vertical strip. (See attached image.)

In the figure, the top of the vertical strip touches the graph of y=x^3. And its lower end touches the graph of y = 3x-2. Also, the bounded region starts at x=-2 and ends at x=1.

Applying the formula

`A = int_a^b (y_(_(upper)) - y_(_(lower)))dx`

the integral needed to compute the area of the bounded region is:

`A = int_(-2)^1 (x^3-(3x-2))dx`

Evaluating the integral, it results to:

`A = int_(-2)^1 (x^3 -3x + 2)dx`

`A = (x^4/4 - (3x^2)/2+2x)` `|_(-2)^1`

`A = (1^4/4 - (3*1^2)/2+2*1) - ((-2)^4/4- (3*(-2)^2)/2+2*(-2))`

`A =3/4-(-6)`

`A=27/4`

**Therefore, the area of the bounded region is `27/4` square units.**