`1/(a-b)+1/(a-2b)+1/(a-3b)+ . . . +1/(a-nb)= alpha.n+betan^2+gamma .n^3 , b/a` is smaller then `(b/a)^3` and other higher powers are neglected . then find `gamma`

If b is very small as compared to the value of a, so that the cube and other higher powers of (b)/(a) can be neglected in the identity (1)/(a-b)+(1)/(a-2b)+(1)/(a-3b)+...+(1)/(a-nb)=alphan+betan^(2)+lambdan^(3) , then the value of lambda is :

If |(alpha^(2n),alpha^(2n+2),alpha^(2n+4)),(beta^(2n),beta^(2n+2),beta^(2n+4)),(gamma^(2n),gamma^(2n+2),gamma^(2n+4))|=((1)/(beta^(2))-(1)/(alpha^(2)))((1)/(gamma^(2))-(1)/(beta^(2)))((1)/(alpha^(2))-(1)/(gamma^(2))) {where alpha^(2), beta^(2) and gamma^(2) are al distinct}, then the value of n is equal to

Show that | (1,1,1), (alpha ^ 2, beta ^ 2, gamma ^ 2), (alpha ^ 3, beta ^ 3, gamma ^ 3) | = (alpha-beta) (beta-gamma) (gamma-alpha) (alphabeta + betagamma + gammaalpha) |

Statement 1: If point (alpha, beta, gamma) lies above the plane (a^(2)+1)x+(b+1)y+(c^(2)+c+1)z+d=0 , then (a^(2)+1)alpha+(b+1)beta+(c^(2)+c+1)gamma+d gt 0 . because Statement 2: If the point (alpha, beta, gamma) lies above the plane ax+by+cz+d=0 then (a alpha+b beta+c gamma+d)/( c )gt 0 .

In crystal system , alpha=beta=90^@,gamma=120^@ and a=1.5,b=1.5,c=3. Find the crystal?

If alpha+beta=gamma and tan gamma=22, a is the arithmetic and b is the geometric mean respectively between tan alpha and tan beta , then the value of (a^3/(1-b^2)^3) is equal to