Subtract the sum of $12 a b - 10 b^{2} - 18 a^{2}$ and $9 a b + 12 b^{2} + 14 a^{2}$ from the sum of $a b + 2 b^{2}$ and $3 b^{2} - a^{2} .$

3 a^{2} - 3 b^{2} - 20 a b

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3 a^{2} + 3 b^{2} - 20 a b

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a^{2} + b^{2} - 20 a b

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3 a^{2} + 3 b^{2} + 20 a b

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Solution

The correct option is B $3 a^{2} + 3 b^{2} - 20 a b$
Sum of $12 a b - 10 b^{2} - 18 a^{2} a n d 9 a b + 12 b^{2} + 14 a^{2}$
$= 12 a b - 10 b^{2} - 18 a^{2} + 9 a b + 12 b^{2} + 14 a^{2}$
On combining the like terms,
$= (12 a b + 9 a b) + (- 10 b^{2} + 12 b^{2}) + (- 18 a^{2} + 14 a^{2})$
Solving the groups,
$= a b (12 + 9) + b^{2} (- 10 + 12) + a^{2} (- 18 + 14)$
$= 21 a b + 2 b^{2} - 4 a^{2}$

Sum of $a b + 2 b^{2}$ and $3 b^{2} - a^{2}$
$= a b + 2 b^{2} + 3 b^{2} - a^{2}$
On combining the like terms,
$= a b + (2 b^{2} + 3 b^{2}) - a^{2}$
$= a b + 5 b^{2} - a^{2}$

Now, subtracting $21 a b + 2 b^{2} - 4 a^{2}$ from $a b + 5 b^{2} - a^{2}$ , we get
$= (a b + 5 b^{2} - a^{2}) - (21 a b + 2 b^{2} - 4 a^{2}) = a b + 5 b^{2} - a^{2} - 21 a b - 2 b^{2} + 4 a^{2}$
On combining the like terms,
$= (a b - 21 a b) + (5 b^{2} - 2 b^{2}) + (- a^{2} + 4 a^{2})$
Solving the groups,
$= - 20 a b + 3 b^{2} + 3 a^{2} = 3 a^{2} + 3 b^{2} - 20 a b$

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Question 71

Subtract the sum of $12 a b - 10 b^{2} - 18 a^{2}$ and $9 a b + 12 b^{2} + 14 a^{2}$ from the sum of $a b + 2 b^{2}$ and $3 b^{2} - a^{2} .$

12 a b - 10 b^{2} - 18 a^{2}

9 a b + 12 b^{2} + 14 a^{2}

a b + 2 b^{2}

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

{9 b}^{2} + {3 c}^{2}

{2 b}^{2} + b c + {2 c}^{2}

{2 b}^{2} - 2 b c - c^{2}

c^{2} + 2 b c - b^{2}

State True or False:

On subtracting $a^{2} + a b + b^{2}$ from $4 a^{2} - 3 a b + 2 b^{2}$ , the answer is $3 a^{2} - 4 a b + b^{2}$ .

a^{3} - 4 a^{2} + 5 a - 6

3 a^{3} + a^{2} + 1

a^{2} - 2

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