non-singular cubic form contains a non-zero term in the shape $x_1^3$ If $f(x_1,...,x_n)$ is a non-singular cubic form over a field of characteristic not equal to 3, why must the expansion of $f(x_1,...,x_n)$ (after an invertible linear transformation of variables) contain a term in the shape $x_1^3$?

Any cubic form $f(x_1,\cdots,x_n)$ may take the form $f(x_1,\cdots,x_n)=ax_1^3+\sum_{i=0}^{2}x_1^i f_{3-i}(x_2,\cdots,x_n)$ where each form $f_j$ is of degree $j.$ One can make a linear transformation to ensure that $f_1\equiv0.$ Then one computes that $\frac{\partial f}{\partial x_i}(1,0,\cdots,0)=0$ for all $2 \leq i \leq n$ and that $\frac{\partial f}{\partial x_1}(1,0,\cdots,0)=a,$ due to $char(k) \neq 3 \Rightarrow x^3 \neq x.$ That is, all partial derivatives ( apart from one ) are zero at the point $(1,0,\cdots,0).$ The form is non-singular, hence the remaining partial derivative should not be zero, i.e. $a \neq 0,$ which what we want. 
