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Question
Find the following product:
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Solution
Given\[\left( \frac{3}{x} - 2 x^2 \right) \left( \frac{9}{x^2} + 4 x^4 - 6x \right)\]
We shall use the identity `(a-b)(a^2 + ab + b^2) = a^3 - b^3`
We can rearrange the \[\left( \frac{3}{x} - 2 x^2 \right) \left( \frac{9}{x^2} + 4 x^4 - 6x \right)\] as
\[\left( \frac{3}{x} - 2 x^2 \right)\left( \left( \frac{3}{x} \right)^2 + \left( 2 x^2 \right)^2 - \left( \frac{3}{x} \right)\left( 2 x^2 \right) \right)\]
\[ = \left( \frac{3}{x} \right)^3 - \left( 2 x^2 \right)^3 \]
\[ = \left( \frac{3}{x} \right)\left( \frac{3}{x} \right)\left( \frac{3}{x} \right) - \left( 2 x^2 \right)\left( 2 x^2 \right)\left( 2 x^2 \right)\]
\[ = \frac{27}{x^3} - 8 x^6\]
Hence the Product value of \[\left( \frac{3}{x} - 2 x^2 \right) \left( \frac{9}{x^2} + 4 x^4 - 6x \right)\] is `27/x^3 - 8x^6`.
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