a^(3)+b^(3)

The value of 2a^(3)-[3a^(3)+4b^(3)-{2a^(3)+(-7a^(3))}5a^(3)-7b^(3)] is (a) -11a^(3)+3b^(3)( b) 7b^(3)+3a^(3)(c)11a^(3)-3b^(3) (d) (11a^(3)+3b^(3))

a^(3)-9b^(3)+(a+b)^(3)

a^(3)+9b^(3)+(a-b)^(3)

If a + b + c = 0 , show that a^(3) + b^(3) + c^(3) = 3abc The following are the steps involved in showing the above result. Arrange them in sequential order (A) a^(3) + b^(3) + 3ab (-c) = -c^(3) (B) (a + b)^(3) = (-c)^(3) (C) a + b + c = 0 rArr a + b = -c (D) a^(3) + b^(3) + 3ab (a +b) = -c^(3) (E) a^(3) + b^(3) + c^(2) = 3abc

If (2a - 3b)/(2a+3b) = 1/3 , then find the value of (2a^(3) - 3b^(3))/(2a^(3) + 3b^(3)) .

((1)/(3)a+(2)/(3)b)^(3)-((1)/(3)a-(2)/(3)b)^(3)

The value of the determinant Delta = |((1 - a_(1)^(3) b_(1)^(3))/(1 - a_(1) b_(1)),(1 - a_(1)^(3) b_(2)^(3))/(1 - a_(1) b_(2)),(1 - a_(1)^(3) b_(3)^(3))/(1 - a_(1) b_(3))),((1 - a_(2)^(3) b_(1)^(3))/(1 - a_(2) b_(1)),(1 - a_(2)^(3) b_(2)^(3))/(1 - a_(2) b_(2)),(1 - a_(2)^(3) b_(3)^(3))/(1 - a_(2) b_(3))),((1 - a_(3)^(3) b_(1)^(3))/(1 - a_(3) b_(1)),(1 - a_(3)^(3) b_(2)^(3))/(1 - a_(3) b_(2)),(1 - a_(3)^(3) b_(3)^(3))/(1 - a_(3) b_(3)))| , is

(2a-3b)^(3)

Factorize: (a-3b)^3+(3b-c)^3+(c-a)^3