Get Brainliest Award For Answering Factorize (x-a)^3+(x-b)^3+(x-c)^3 where x= [(a+b+c)/(3)]
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as we know
given x=a+b+c/3
a+b+c=3x
We know that,
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
If a+b+c=0 then,
a³+b³+c³-3abc=0×(a²+b²+c²-ab-bc-ca)=0 .....(1)
(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)
let a=(x-a)
b=(x-b)
and c=(x-c)
a+b+c=x-a+x-b+x-c=3x-(a+b+c)=3x-3x=0....(as a+b+c=3x)
(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)=0
Thus the answer is 0....
given x=a+b+c/3
a+b+c=3x
We know that,
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
If a+b+c=0 then,
a³+b³+c³-3abc=0×(a²+b²+c²-ab-bc-ca)=0 .....(1)
(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)
let a=(x-a)
b=(x-b)
and c=(x-c)
a+b+c=x-a+x-b+x-c=3x-(a+b+c)=3x-3x=0....(as a+b+c=3x)
(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)=0
Thus the answer is 0....
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