Zeroes of $27 - x^{3} - 9 x^{2} + 27 x$ are/is

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-3 and 3

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1, -3 and 3

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0, 3 and -3

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Solution

The correct option is A 3
Given, $27 - x^{3} + 9 x^{2} - 27 x$

We can write it as, $27 - x^{3} - 27 x + 9 x^{2} = 3^{3} - x^{3} - 9 x (3 - x) 3^{3} - x^{3} - 3 \times 3 x (3 - x) (a - b)^{3} = a^{3} - b^{3} - 3 a b (a - b) 3^{3} - x^{3} - 9 x (3 - x) = (3 - x)^{3}$

Lets find zeroes now, $(3 - x)^{3} = 0 \Rightarrow 3 - x = 0$
We get $x = 3$

The zero is 3

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