The product of $4 a^{2}, - 6 b^{2}$ and $3 a^{2} b^{2}$ is

a^{2} b^{2}

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13 a^{4} b^{4}

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- 72 a^{4} b^{4}

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a^{4} b^{4}

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Solution

The correct option is C $- 72 a^{4} b^{4}$
$(4 a^{2}) (- 6 b^{2}) (3 a^{2} b^{2})$
= $4 \times (- 6) \times 3 \times a^{2 + 2} . b^{2 + 2}$
= $- 72 a^{4} b^{4}$

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Similar questions

4 a^{2} - 6 b^{2} + 8 a b - 2 a^{2} + 8 b^{2} + 13 a b

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= 5

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a^{2} b^{2} + 4 a^{2}

(4 a^{2} + 2 b) (a^{2} b + a b)

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