Question

Find the product of [3d^2+c^2] and [2d^2-3c^2]

Answer 1

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Solution :-

$= > ( {3d}^{2} + {c}^{2} ) \times ( {2d}^{2} - {3c}^{2} )$

$= > {6d}^{2} - {4c}^{2}$

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Answer 2

Answer:

6d² - 4c²

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