Math, asked by rkcomp31, 1 month ago

If a+b+c=0 then prove that a³+b³+c³=3abc.

adarsharyan46: only reverse it

adarsharyan46: start with my last steps

adarsharyan46: also i cannot answer that cuz aleready 2 people answered

adarsharyan46: ;)

adarsharyan46: also not completely reverse it, i mean at some points you need to use the given statement

adarsharyan46: and proove it

adarsharyan46: ok i have answered it

adarsharyan46: cuz i guess a moderator deleted one of my question

adarsharyan46: i mean one of the answers

adarsharyan46: that was inappropriate

Answers

Answered by bson

Step-by-step explanation:

a+b+c =0

a+b = -c

(a+b)³ =-c³

a³+b³+3a²b+3ab² = -c³

a³+b³+c³+3ab(a+b)=0

a³+b³+c³+3ab*(-c)=0

a³+b³+c³-3abc = 0

a³+b³+c³ = 3abc

Answered by adarsharyan46

Step-by-step explanation:

We have :

a + b + c = 0

⇒ a + b = -c --------- (i)

now,

Cubing both sides, we get

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

⇒ $a^{3}+ b^{3} + 3ab(a+ b ) = - c^{3}$

⇒ $a^{3}+ b^{3} + 3ab(-c ) = - c^{3}$ [ as a + b = -c from (i) ]

⇒ $a^{3}+ b^{3} - 3abc = - c^{3}$

Now take $-3abc$ to RHS and $-c^{3}$ to LHS

⇒ $a^{3}+ b^{3} + c^{3} = 3abc$

hence proved..

HOPE IT HELPS

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