If the centroid of the triangle formed by the points (a,b),(b,c) and (c,a) is at the origin, then a 3 +b 3 +c 3 .
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Answered by
272
Coordiante of the centroid of the triangle is:
x=(x₁+x₂+x₃)/3 , y=(y₁+y₂+y₃)/3
∴, According to the given question,
(a+b+c)/3=0
or, a+b+c=0 -----------------(1)
Now, (a+b+c)³
=a³+3a²b+3ab²+b³+3a²c+6abc+3b²c+3ac²+3bc²+c³
or, (a+b+c)³=a³+b³+c³+3a²(b+c)+3b²(c+a)+3c²(a+b)+6abc
or, a³+b³+c³=(a+b+c)³-3a²(-a)-3b²(-b)-3c²(-c)-6abc [Using (1)]
or, a³+b³+c³=3a³+3b³+3c³-6abc
or, a³+b³+c³-3a³-3b³-3c³=-6abc
or, -2a³-2b³-2c³=-6abc
or, a³+b³+c³=3abc Ans.
x=(x₁+x₂+x₃)/3 , y=(y₁+y₂+y₃)/3
∴, According to the given question,
(a+b+c)/3=0
or, a+b+c=0 -----------------(1)
Now, (a+b+c)³
=a³+3a²b+3ab²+b³+3a²c+6abc+3b²c+3ac²+3bc²+c³
or, (a+b+c)³=a³+b³+c³+3a²(b+c)+3b²(c+a)+3c²(a+b)+6abc
or, a³+b³+c³=(a+b+c)³-3a²(-a)-3b²(-b)-3c²(-c)-6abc [Using (1)]
or, a³+b³+c³=3a³+3b³+3c³-6abc
or, a³+b³+c³-3a³-3b³-3c³=-6abc
or, -2a³-2b³-2c³=-6abc
or, a³+b³+c³=3abc Ans.
kvnmurty:
we could use the rule/law that if a +b+c =0, then a^3+b^3+c^3 = 3 a b c... directly..
Answered by
20
Step-by-step explanation:
Coordiante of the centroid of the triangle is:
x=(x₁+x₂+x₃)/3 , y=(y₁+y₂+y₃)/3
∴, According to the given question,
(a+b+c)/3=0
or, a+b+c=0 -----------------(1)
Now, (a+b+c)³
=a³+3a²b+3ab²+b³+3a²c+6abc+3b²c+3ac²+3bc²+c³
or, (a+b+c)³=a³+b³+c³+3a²(b+c)+3b²(c+a)+3c²(a+b)+6abc
or, a³+b³+c³=(a+b+c)³-3a²(-a)-3b²(-b)-3c²(-c)-6abc [Using (1)]
or, a³+b³+c³=3a³+3b³+3c³-6abc
or, a³+b³+c³-3a³-3b³-3c³=-6abc
or, -2a³-2b³-2c³=-6abc
or, a³+b³+c³=3abc Ans.
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