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
shadowsabers03:
0 is the answer.
Answers
Answered by
2
Answer: 0
Identity
a^3 + b^3+c^3-3abc
=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
=(0)(......................)
=0
Hope it helps
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.
Answered by
6
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!
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