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

$(-2a)+(-2b)+(-2c) \\ \\ -2a-2b-2c \\ \\ -2(a+b+c) \\ \\ -2 \times 0 \ = \ 0$

$\sf{Let}$

$-2a=x \\ \\ -2b=y \\ \\ -2c=z$

$\therefore\ (-2a)+(-2b)+(-2c)=x+y+z=0$

$(-2a)^3 + (-2b)^3 + (-2c)^3 - 3 (-2a)(-2b)(-2c) \\ \\ \\ x^3+y^3+z^3-3xyz \\ \\ \\ (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \\ \\ \\ 0(x^2+y^2+z^2-xy-yz-zx) \\ \\ \\ 0$

$\sf{Hence proved! \\ \\ \\ Thank you. :-)}$