The factors of (2a-b³)+(b-2c)³+(b-2a)³+8(c-a)³ is
Answers
Answered by
10
Answer:
The given equation is,
(2a - b)³ + (b - 2c)³ + 8(c - a)³
= (2a - b)³ + (b - 2c)³ + (2c - 2a)³
Let x = (2a - b) ; y = (b - 2c) ; z = (2c - 2a)
Now,
x + y+ z = (2a - b) + (b - 2c) + (2c - 2a) = 0
Since, x + y + z = 0, then,
x³ + y³ + z³ = 3xyz
⇒ (2a - b)³ + (b - 2c)³ + (2c - 2a)³ = 3(2a - b) (b - 2c) (2c - 2a)
⇒ (2a - b)³ + (b - 2c)³ + 8(c - a)³ = 6(2a - b) (b - 2c) (c - a)
here is the correct answer please mark me brainliest and give thanks
Similar questions