Question

the factors of (2a-b)³+(b-2c)³+8(c-a)³​

Answer 1

Step-by-step explanation:

Given :-

(2a-b)³+(b-2c)³+8(c-a)³

To find :-

Find the factors of (2a-b)³+(b-2c)³+8(c-a)³?

Solution :-

Given expression is (2a-b)³+(b-2c)³+8(c-a)³

It can be written as (2a-b)³+(b-2c)³+2³(c-a)³

=> (2a-b)³+(b-2c)³+[2(c-a)]³

=> (2a-b)³+(b-2c)³+(2c-2a)³

This is in the form of x³+y³+z³

Where , x = 2a-b , y = b-2c and z = 2c-2a

and

x+y+z = 2a-b+b-2c+2c-2a

=> x+y+z = (2a-2a)+(b-b)+(2c-2c)

=> x+y+z = 0

We know that

If x+y+z = 0 then x³+y³+z³ = 3xyz

Now,

We have

(2a-b)³+(b-2c)³+(2c-2a)³

=> 3(2a-b)(b-2c)(2c-2a)

=> 3(2a-b)(b-2c)(2)(c-a)

=> 6(2a-b)(b-2c)(c-a)

The factorization of the given expression is 6(2a-b)(b-2c)(c-a)

Answer :-

The factors of the given expression are 1, 6, (2a-b) , (b-2c) and (c-a)

Used formulae:-

• (ab)^m = a^m × a^n
• If x+y+z = 0 then x³+y³+z³ = 3xyz
• 1 is a factor of every number.