Find out how much does $a^{4} - 3 a^{2} b^{2} + b^{4}$ exceed $3 a^{4} - 2 a^{2} b^{2} + b^{4} + 3 a^{2} b + 2 a b^{2} .$

(- 2 a^{4} - a^{2} b^{2} + 3 a^{2} b - 2 a b^{2})

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- (2 a^{4} - a^{2} b^{2} - 3 a^{2} b - 2 a b^{2})

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

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(- 2 a^{4} - a^{2} b^{2} - 3 a^{2} b - 2 a b^{2})

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Solution

The correct option is D $(- 2 a^{4} - a^{2} b^{2} - 3 a^{2} b - 2 a b^{2})$
To find out the resulting expression, we have to subtract $(3 a^{4} - 2 a^{2} b^{2} + b^{4} + 3 a^{2} b + 2 a b^{2})$ from $(a^{4} - 3 a^{2} b^{2} + b^{4})$ as shown below:

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

Collecting positive and negative like terms together, we get:

$= a^{4} - 3 a^{4} - 3 a^{2} b^{2} + 2 a^{2} b^{2} + b^{4} - b^{4} - 3 a^{2} b - 2 a b^{2} = - 2 a^{4} - a^{2} b^{2} - 3 a^{2} b - 2 a b^{2}$

Hence, the resulting expression is $(- 2 a^{4} - a^{2} b^{2} - 3 a^{2} b - 2 a b^{2})$ .

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