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Answer :
Let \(p(x) = x^3 - ax^2 + 6x - a\)
The zero of \(x - a\) is x = a
[\(\because (x - a = 0) \) ]
\(\therefore p(a) = (a)^3 - a(a)^2 + 6(a) - a\)
\(\Rightarrow p(a) = a^3 - a^3 + 6a - a\)
\(\Rightarrow p(a) = 5a\)
Hence, By remainder theorem, required remainder = 5a.