To find the equation of the tangent to the curve $x^3 + 2xy^2 + y^3 - 21 = 0$ at the point $(1, 2)$, we will follow these steps:

1. Differentiate the equation implicitly:

We implicitly differentiate both sides of the given equation with respect to $x$, remembering that $y$ is a function of $x$.

Given equation: $x^3 + 2xy^2 + y^3 - 21 = 0$

Now, let's differentiate term by term:

1. $\frac{d}{dx}(x^3) = 3x^2$
2. $\frac{d}{dx}(2xy^2) = 2 \left( \frac{d}{dx}(x) \cdot y^2 + x \cdot \frac{d}{dx}(y^2) \right) = 2(y^2 + x \cdot 2y \cdot \frac{dy}{dx})$
3. $\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$
4. $\frac{d}{dx}(-21) = 0$

Putting everything together:

$3x^2 + 2(y^2 + 2xy \cdot \frac{dy}{dx}) + 3y^2 \cdot \frac{dy}{dx} = 0$

Simplifying:

$3x^2 + 2y^2 + 4xy \cdot \frac{dy}{dx} + 3y^2 \cdot \frac{dy}{dx} = 0$

Combine like terms:

$3x^2 + 2y^2 + (4xy + 3y^2) \cdot \frac{dy}{dx} = 0$

Now solve for $\frac{dy}{dx}$:

$(4xy + 3y^2) \cdot \frac{dy}{dx} = -3x^2 - 2y^2$

$\frac{dy}{dx} = \frac{-3x^2 - 2y^2}{4xy + 3y^2}$

2. Find the slope of the tangent at the point $(1, 2)$:

Substitute $x = 1$ and $y = 2$ into the derivative:

$\frac{dy}{dx} = \frac{-3(1)^2 - 2(2)^2}{4(1)(2) + 3(2)^2}$

$\frac{dy}{dx} = \frac{-3 - 8}{8 + 12}$

$\frac{dy}{dx} = \frac{-11}{20}$

So, the slope of the tangent line at the point $(1, 2)$ is $-\frac{11}{20}$.

3. Use the point-slope form to find the equation of the tangent line:

The point-slope form of the equation of a line is:

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

Where $m$ is the slope, and $(x_1, y_1)$ is the given point.

Using $m = -\frac{11}{20}$, $x_1 = 1$, and $y_1 = 2$:

$y - 2 = -\frac{11}{20}(x - 1)$

Simplifying:

$y - 2 = -\frac{11}{20}x + \frac{11}{20}$

$y = -\frac{11}{20}x + \frac{11}{20} + 2$

$y = -\frac{11}{20}x + \frac{11}{20} + \frac{40}{20}$

$y = -\frac{11}{20}x + \frac{51}{20}$

Final Answer:

The equation of the tangent line is:

$y = -\frac{11}{20}x + \frac{51}{20}$

Do you want more details or have any questions? Here are some related questions to explore:

1. How do you find the normal line to a curve at a given point?
2. Can we find the tangent to a parametric curve?
3. What are the applications of implicit differentiation in real life?
4. How do tangents help in approximation using linearization?
5. How can we determine concavity and inflection points using the second derivative?

Tip: Always keep track of your derivative rules when using implicit differentiation, especially with product and chain rules.