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The answer to the third question is six 6. By using the identity.
PRACHIsheoran:
okk
then
Then it will become 2(a+b+c)^2
Then use the identity( a+b+c)^2
Okay
okkk thanks
Mark it
Brainliest
i don't know how to mark it as brainlist
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Answer :Option (d)
a+b+c = 0
a = -(b+c) OR b = -(a+c) OR c = -(b+c)
Now substitute this values in 1st, 2nd , 3rd term respectively:
1/b^2+c^2-a^2 = 1/b^2+c^2-(-(b+c))^2 = 1/b^2+c^2-b^2-c^2-2bc = 1/-2bc = -1/2bc
1/c^2+a^2-b^2 = 1/c^2+a^2-(-(a+c))^2 = -1/2ac
1/a^2+b^2-c^2 = 1/a^2+b^2-(-(b+c))^2 = -1/2ab
So now the equation reads:
-1/2bc + -1/2ac + -1/2ab = -1/2 *(1/bc + 1/ac + 1/ab) = -1/2 * (a+b+c/abc)
= -1/2 * (0/abc) = 0... [a+b+c = 0, given]
Hence answer = 0.
Alternatively:
a+b+c = 0
Hence substitute: a= 1/2 , b= 1/2 , c = -1
1/(1/4+1-1/4) + 1/(1+1/4-1/4) + 1/(1/4+1/4-1)
1 + 1 + 1/(1/2 - 1) = 1 + 1 + 1/(1-2)/2 = 1 + 1 + 2/-1 = 1+1-2 =0
Hope it helps
a+b+c = 0
a = -(b+c) OR b = -(a+c) OR c = -(b+c)
Now substitute this values in 1st, 2nd , 3rd term respectively:
1/b^2+c^2-a^2 = 1/b^2+c^2-(-(b+c))^2 = 1/b^2+c^2-b^2-c^2-2bc = 1/-2bc = -1/2bc
1/c^2+a^2-b^2 = 1/c^2+a^2-(-(a+c))^2 = -1/2ac
1/a^2+b^2-c^2 = 1/a^2+b^2-(-(b+c))^2 = -1/2ab
So now the equation reads:
-1/2bc + -1/2ac + -1/2ab = -1/2 *(1/bc + 1/ac + 1/ab) = -1/2 * (a+b+c/abc)
= -1/2 * (0/abc) = 0... [a+b+c = 0, given]
Hence answer = 0.
Alternatively:
a+b+c = 0
Hence substitute: a= 1/2 , b= 1/2 , c = -1
1/(1/4+1-1/4) + 1/(1+1/4-1/4) + 1/(1/4+1/4-1)
1 + 1 + 1/(1/2 - 1) = 1 + 1 + 1/(1-2)/2 = 1 + 1 + 2/-1 = 1+1-2 =0
Hope it helps
Sorry,
Your answer is right and brainliest also but i marked other answer as brainliest
by mistake
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