a cube + b cube + c cube =

a cube + b cube + c cube - 3 abc = (a+b+c) (a sq.+ b sq + c sq -ab-bc-ca)
In the sums based on this identity if LHS is given for factorisation and LHS has 3 terms (instead of 4 terms as seen above) i.e. 2 terms are perfect cube and 3rd term is not a perfect cube and has coefficient D then;
(I) first we should find the cube roots of the 2 perfect cubes,
(II) and substitute them (those) in the places of a & b above,
(III) compare the 3rd non perfect cube term with c cube - 3abc
(IV) Since only like terms can be added or subtracted, therefore you will get what should be variable part in 3rd perfect cube ( if it has so) . Let m be the coefficient part of the 3rd perfect cube
(V) and problem reduces to the form m cube + Cm = D
Here C = (-3) (coefficient part of a) (coefficient part of b)
repeating, D = coefficient part of the non-perfect cube term in the question given to solve.

m3+Cm=D

STEPS:

for small values of root of m;-

1) Find factors of D

2) Take bigger factor first say q (first FACTOR should be D itself)

3) Find (q-C) till it is a perfect square say r square

4) Divide D by this factor q to get solution r i.e. the required value of m that satisfies the above equation.

If you square this root; you will get the perfect square you got in step 3)

In other words, That factor of D is root of the above equation in variable m for which: square of answer in step 4 = answer in step 3 5) Remember factors include -ve integers also. Ex. -3 is also a factor of 15. (VI) Once you get m you get c = m x variable part obtained in step (IV)

(VII) Substitute  a, b & c in the RHS of the identity at the top , to get required factors. Simplify by clubbing like terms.