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

We know that=>

a+b+c = 0

To find=>

(-2a)^3 + (-2b)^3+(-2c)^3 -3(-2a)(-2b)(-2c)

Now, We have

(a+b+c)^3

=>a^3+b^3+c^3 + 3(a+b)(b+c)(a+c).

By using this identity we put=>

(-2a)^3 + (-2b)^3+(-2c)^3 -3(-2a)(-2b)(-2c)

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

=>{-2(a+b+c)}^3

(Taking -2 common from the terms)

=>{-2(0)}^3 (Since a+b+c = 0)

=>0

(since anything multiplied by 0 is 0)

Hence,the value of (-2a)^3 + (-2b)^3 + (-2c)^3 - 3 (-2a)(-2b)(-2c) will be 0.

Remember

The identity used here=>

(a+b+c)^3

=>a^3+b^3+c^3 + 3(a+b)(b+c)(a+c).

Also, remember

(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2