<s>[INST] Exercise on the following surface $\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}=\sqrt[4]{c^3}$ for $c\ge 0$. Consider the surface $\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}=\sqrt[4]{c^3}$ for $c\ge 0$. Find the plane tange to the surface at the point $ (x_0, y_0, z_0) $ and show that the sum of the cubes of the intersections of said plane with the axes is equal to one constant.
Try: To find the equation of the tangent plane, let's first find the derivatives. Let $F (x, y, z) = x^{ 3/4} + y ^{3/4} + z^{ 3/4}$. Then,
$$ F_x (x_0, y_0, z_0) = \dfrac {3} {4x_0 ^ {1/4}}, \quad F_y (x_0, y_0, z_0) = \dfrac {3} {4y_0 ^ {1/4} }, \quad F_z (x_0, y_0, z_0) = \dfrac {3} {4z_0 ^ {1/4}}. $$
$$\begin{align*}\dfrac {3} {4x_0 ^ {1/4}}(x-x_0)+\dfrac {3} {4y_0 ^ {1/4} }(y-y_0)+\dfrac {3} {4z_0 ^ {1/4}}(z-z_0)&=0\\ \dfrac{3}{4}\left (\dfrac {1} {x_0 ^ {1/4}}(x-x_0)+\dfrac {1} {y_0 ^ {1/4} }(y-y_0)+\dfrac {1} {z_0 ^ {1/4}}(z-z_0) \right )&=0\\\dfrac {1} {x_0 ^ {1/4}}(x-x_0)+\dfrac {1} {y_0 ^ {1/4} }(y-y_0)+\dfrac {1} {z_0 ^ {1/4}}(z-z_0) &=0  \end{align*}$$
Can someone help me with the other part of the question and tell me if my tangent plane is ok. Thanks a lot.
 [/INST] You can rewrite the equation of the tangent plane as,
$\dfrac {x} {x_0 ^{1/4}} + \dfrac {y} {y_0 ^ {1/4} } + \dfrac {z} {z_0 ^ {1/4}} = x_0 ^{3/4} + y_0 ^{3/4} + z_0 ^{3/4} = c^{3/4} \tag1$
Now at the intersection of the tangent plane with x-axis, $y = 0,  z = 0$.
So plugging into $(1)$,
$x = x_0 ^{1/4} c^{3/4}$
Similarly at intersection with other axes,
with y-axis: $y = y_0 ^{1/4} c^{3/4}$
with z-axis: $z = z_0 ^{1/4} c^{3/4}$
Now using $ \ x_0 ^{3/4} + y_0 ^{3/4} + z_0 ^{3/4} = c^{3/4} , \ $ you can show that the sum of cubes of x, y and z intercepts is a constant $( = c^3)$.
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