Q1. Find the value of (x-a)3+(x-b)3+(x-c)3-3(x-a)(x-b)(x-c), if a+b+c=3x.
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Solution
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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