Answer
Using Theorem 5 of Section 15.5, we show that the tangent plane to the surface ${x^n} + {y^n} + {z^n} = r$ at $P = \left( {a,b,c} \right)$ has equation
${a^{n - 1}}x + {b^{n - 1}}y + {c^{n - 1}}z = r$
Work Step by Step
Write the surface:
$F\left( {x,y,z} \right) = {x^n} + {y^n} + {z^n} = r$
The partial derivatives are
${F_x} = n{x^{n - 1}}$, ${\ \ \ }$ ${F_y} = n{y^{n - 1}}$, ${\ \ \ }$ ${F_z} = n{z^{n - 1}}$
At $P = \left( {a,b,c} \right)$,
${F_x}\left( {a,b,c} \right) = n{a^{n - 1}}$, ${\ \ \ }$ ${F_y}\left( {a,b,c} \right) = n{b^{n - 1}}$, ${\ \ \ }$ ${F_z}\left( {a,b,c} \right) = n{c^{n - 1}}$
By Theorem 5 of Section 15.5, the tangent plane to the surface at $P = \left( {a,b,c} \right)$ has equation
${F_x}\left( {a,b,c} \right)\left( {x - a} \right) + {F_y}\left( {a,b,c} \right)\left( {y - b} \right) + {F_z}\left( {a,b,c} \right)\left( {z - c} \right) = 0$
So,
$n{a^{n - 1}}\left( {x - a} \right) + n{b^{n - 1}}\left( {y - b} \right) + n{c^{n - 1}}\left( {z - c} \right) = 0$
$n{a^{n - 1}}x - n{a^n} + n{b^{n - 1}}y - n{b^n} + n{c^{n - 1}}z - n{c^n} = 0$
Dividing both sides by $n$ and re-arranging gives
${a^{n - 1}}x + {b^{n - 1}}y + {c^{n - 1}}z - \left( {{a^n} + {b^n} + {c^n}} \right) = 0$
Since $P = \left( {a,b,c} \right)$ is a point on the surface, we have
${a^n} + {b^n} + {c^n} = r$
So,
${a^{n - 1}}x + {b^{n - 1}}y + {c^{n - 1}}z - r = 0$
${a^{n - 1}}x + {b^{n - 1}}y + {c^{n - 1}}z = r$
Hence, the tangent plane to the surface ${x^n} + {y^n} + {z^n} = r$ at $P = \left( {a,b,c} \right)$ has equation
${a^{n - 1}}x + {b^{n - 1}}y + {c^{n - 1}}z = r$
