If `a+b+c=9` and `a b+b c+c a=23` , then `a^3+b^3+c^3-3a b c=?` (a)108 (b) 207 (c) 669 (d) 729

If a+b+c=9 and ab+bc+ca=23 then a^(3)+b^(3)+c^(3)-3abc=108(b)207(c)669(d)729

If a+b+c=6\ a n d\ a b+b c+c a=11 , find the value of a^3+b^3+c^3-3a b c

If a+b+c=9 and a^2+b^2+c^2=35 , find the value of a^3+b^3+c^3-3a b c

If a+b+c=9 and a^2+b^2+c^2=35 , find the value of a^3+b^3+c^3-3a b c

If a+b+c=9\ \ \ a n d\ \ \ a b+b c+c a=26 , find the value of a^3+b^3+c^3-3a b c

If the centroid of the triangle formed by the points (a ,\ b),\ (b ,\ c) and (c ,\ a) is at the origin, then a^3+b^3+c^3= a b c (b) 0 (c) a+b+c (d) 3\ a b c

If a^3 - b^3 - c^3 = 0 then the value of a^9 - b^9 - c^9 -3a^3 b^3 c^3 is

Prove: \ |(a, b, c),( a-b,b-c,c-a),( b+c,c+a, a+b)|=a^3+b^3+c^3-3a b c

If a+b+c=0 , then prove a^3+b^3+c^3=3a(c+a)(b+a)=3b(b+c)(b+a)=3c(c+a)(c+b)

Prove that (a+b+c)^3 - a^3-b^3-c^3 = 3(a+b)(b+c)(c+a).