[" Give me this type of question "],[a(a-2b-c)+2bc]

Factorise : a(a-2b-c)+2bc

Long-answer type questions (L.A.) (a)/(x-a)+(b)/(x-b)=(2c)/(x-c)(xnea,b,c) .

Prove that the following. [[a,b,c],[a^2,b^2,c^2],[bc,ca,ab]] =(b-c)(c-a)(a-b)(bc+ca+ab)

If ab + bc + ca = 0, then what is the value of (a^(2))/(a^(2) - bc) + (b^(2))/(b^(2)-ca) + (c^(2))/(c^(2) - ab) ?

The value of the determinant ,a+b+2c,a,bc,b+c+2a,bc,a,c+a+2b]| is

If a,b, and c are in H.P.then th value of ((ac+ab-bc)(ab+bc-ac))/((abc)^(2)) is ((a+c)(3a-c))/(4a^(2)c^(2)) b.(2)/(bc)-(1)/(b^(2)) c.(2)/(bc)-(1)/(a^(2)) d.((a-c)(3a+c))/(4a^(2)c^(2))

Prove the following: [[-a^2,ab,ac],[ab,-b^2,bc],[ac,bc,-c^2]]=4a^2b^2c^2