Expand the algebraic expression $(3 a^{2} + 2 b^{3}) (3 a^{2} - 2 b^{3}) .$

9 a^{4} - 4 b^{9}

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9 a^{4} - 4 b^{6}

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3 a^{4} - 2 b^{6}

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3 a^{4} - 2 b^{9}

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Solution

The correct option is B $9 a^{4} - 4 b^{6}$
$Need to simplify: - ------------------- - (3 a^{2} + 2 b^{3}) (3 a^{2} - 2 b^{3})$

$(3 a^{2} + 2 b^{3}) \times (3 a^{2} - 2 b^{3})$

Applying distributive property of multiplication: $(a + b) \times c = a \times c + b \times c$

$= 3 a^{2} (3 a^{2} - 2 b^{3}) + 2 b^{3} (3 a^{2} - 2 b^{3})$

$= 3 a^{2} \cdot 3 a^{2} - 3 a^{2} \cdot 2 b^{3} + 2 b^{3} \cdot 3 a^{2} - 2 b^{3} \cdot 2 b^{3}$

$= 3 \cdot 3 \cdot a^{2} \cdot a^{2} - 3 \cdot 2 \cdot a^{2} \cdot b^{3} + 2 \cdot 3 \cdot b^{3} \cdot a^{2} - 2 \cdot 2 \cdot b^{3} \cdot b^{3}$

$= 9 a^{2 + 2} - 6 a^{2} b^{3} + 6 a^{2} b^{3} - 4 b^{3 + 3}$
$[∵ p^{m} \cdot p^{n} = p^{m + n}]$

$= 9 a^{4} - 4 b^{6}$
$[∵ - 6 a^{2} b^{3} and + 6 a^{2} b^{3}$ cancel out each other $]$

$∴ (3 a^{2} + 2 b^{3}) (3 a^{2} - 2 b^{3}) = 9 a^{4} - 4 b^{6} - ----------------------------------- -$

Hence, option (b.) is the correct one.

$Note: - ----- -$
We get:
$(3 a^{2} + 2 b^{3}) (3 a^{2} - 2 b^{3})$
$= 9 a^{4} - 4 b^{6}$
$= (3 a^{2})^{2} - (2 b^{3})^{2}$
$\Rightarrow (3 a^{2} + 2 b^{3}) (3 a^{2} - 2 b^{3}) = (3 a^{2})^{2} - (2 b^{3})^{2}$
Hence, we can generalize: $(p + q) (p - q) = p^{2} - q^{2}$

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