By how much is $6 a^{2} b^{2} - 19 a b^{2} - 11 a^{2} b$ greater than $- 6 a^{2} + 7 a^{2} b - 8 a b$ ?

h+r

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All of the above

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Solution

The correct option is B
h

The required result is,

$(6 a^{2} b^{2} - 19 a b^{2} - 11 a^{2} b) - (- 6 a^{2} + 7 a^{2} b - 8 a b)$

= $6 a^{2} b^{2} - 19 a b^{2} - 11 a^{2} b + 6 a^{2} - 7 a^{2} b + 8 a b$

= $6 a^{2} b^{2} - 19 a b^{2} - 11 a^{2} b - 7 a^{2} b + 6 a^{2} + 8 a b$

= $6 a^{2} b^{2} - 19 a b^{2} - 18 a^{2} b + 6 a^{2} + 8 a b$

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