Math, asked by pubgg54, 1 year ago

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​

Answers

Answered by mathsdude85
7

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!

Answered by Anonymous
9
If a + b + c = 0, then

a = - ( b + c ), b = - ( a + c ) and c = - ( a + b )

Replacing the values in the question we obtain,

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

Now taking L.C.M

a^3 + b^3 + c^3 /abc

Using the identity,

If a +b +c = 0, then a^3 + b^3 + c^3 = 3abc

Hence, we get

3abc/abc = 3

Therefore , the value is 3.

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