Question

without finding the cubes factorise ( 7a-3b)^3 + (3b-2c)^3+(2c-7a)^3

Answer 1

Answer:

3(7a - 3b)(3b - 2c)(2c - 7a)

Step-by-step explanation:

Let X = (7a - 3b)

Y = 3b - 2c

Z = 2c - 7a

Here you will observe that

X + Y + Z = 7a - 3b + 3b - 2c + 2c - 7a = 0

We know that, if X + Y + Z = 0 then

X³ + Y³ + Z³ = 3 X Y Z

∴ ( 7a - 3b)³ +  (3b - 2c)³ + ( 2c - 7a)³ = 3(7a - 3b)(3b - 2c)(2c - 7a)