a^3 + b^3 = ?

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

Add: a + b – 3, b – a + 3, a – b + 3

Add: a + b - 3, b - a + 3, a - 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))

Factroise (2a+3b)^3-(2a-3b)^3 .

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

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

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