How to show that the deltoid is a plane algebraic curve of degree $4$? 
How can I show that a deltoid is a plane algebraic curve of degree 4?

I have searched that the parametric equation for deltoid is given by
$$\begin{align}
x&= 2 \cos t + \cos 2t \\
y&=2 \sin t - \sin 2t
\end{align}$$
So, by using some trigonometric properties, we can write
$$\begin{align}
x = 2 \cos t + 2 \cos^2 t - 1  &\quad\to\quad x^2 = \phantom{-}4 \cos^4t + 8\cos^3 t - 4 \cos t + 1 \\
y =2 \sin t -2\sin t \cos t\;\, &\quad\to\quad y^2=-4 \cos^4t+ 8 \cos^3t -8\cos t+4 
\end{align}$$

Write the equations as
$
x=2c^2+2c-1\\
y^2=-4c^4+8c^3-8c+4
$
where $c=\cos t$.
Feeding
Eliminate[{x==2c+2c2-1,y2=-4c^4+8c^3-8c+4},c]

to WA
gives
$$
x^4 - 8 x^3 + x^2 (2 y^2 + 18) + 24 x y^2 = -y^4 - 18 y^2 + 27
$$
You can do this by hand by computing the resultant of $2c^2+2c-1-x$ and $-4c^4+ 8 c^3 -8c+4-y^2$, seen as polynomials in $c$.
The resultant is the determinant of this $6 \times 6$ Sylvester matrix:
$$
\pmatrix{
              2 & 2 & -x-1 & 0 & 0 & 0 \\
              0 & 2 & 2 & -x-1 & 0 & 0 \\
              0 & 0 & 2 & 2 & -x-1 & 0 \\
              0 & 0 & 0 & 2 & 2 & -x-1 \\
              -4 & 8 & 0 & -8 & 4-y^2 & 0 \\
              0 & -4 & 8 & 0 & -8 & 4-y^2 \\
}
$$
It will take some work to compute this determinant by hand but it's not impossible.
However, you don't need to compute the determinant to prove that the curve is algebraic, because the determinant is clearly a polynomial in $x$ and $y$. It is also clear that its is degree $4$.
