Question

If a+b+c=0 , then find the value of ..... (-2a)^3 + (-2b)^3 + (-2c)^3 - 3 (-2a)(-2b)(-2c)............. plzzzz answer me fast I will mark your answer as brainlist​

Answer 1

Answer:

0

Step-by-step explanation:

Given Equation is (-2a)³ + (-2b)³ + (-2c)³ - 3(-2a)(-2b)(-2c)

Now,

∴ a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

⇒ (-2a)³ + (-2b)³ + (-2c)³ - 3(-2a)(-2b)(-2c) = (0)[a² + b² + c² - ab - bc - ca]

⇒ (-2a)³ + (-2b)³ + (-2c)³ = 0.

Hope it helps!

Answer 2

Answer:

Step-by-step explanation:

Answer: 0

Identity

a^3 + b^3+c^3-3abc

=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

=(0)(......................)

=0

Bro., you can take like x = -2a, y = -2b and z= -2c, and then apply x, y and z in your answer. Thus you get the proof. Thanks.