Question

Find the following product:

\[\left( \frac{3}{x} - 2 x^2 \right) \left( \frac{9}{x^2} + 4 x^4 - 6x \right)\]

Answer 1

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`.