On the number of zeros of diagonal cubic forms

Let M<SUB>s</SUB>, be the number of solutions of the equation X<SUB>1</SUB><SUP>3</SUP>+ X<SUB>2</SUB><SUP>3</SUP>+ &#8230; + X<SUB>s</SUB><SUP>3</SUP>=0 in the finite field GF(p). For a prime p &#8801; 1(mod 3), &#8721;<SUP>&#8734;</SUP><SUB>s=1</SUB> M<SUB>s</SUB>X<SUP>3</SUP> = (x/1-px)+((x<SUP>2</SUP>(p-1)(2+dx))/(1-3px<SUP>2</SUP>-pdx<SUP>3</SUP>)), M<SUB>3</SUB>= p<SUP>2</SUP> + d(p - 1), and M<SUB>4</SUB> = p<SUP>2</SUP> + 6(p<SUP>2</SUP> &#8722; p). Here d is uniquely determined by 4<SUB>p</SUB> = d<SUP>2</SUP> + 27b<SUP>2</SUP> and d &#8801; 1(mod 3)