Simply using algebraic identities and match the following.

9 p^{2} + 9 q^{2} + 18 p q

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999900

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

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Solution

Case 1 :
Using $(a + b)^{2} = a^{2} + b^{2} + 2 a b$
$a = 3 p, b = 3 q$

$(3 p + 3 q)^{2} = (3 p)^{2} + (3 q)^{2} + 2 \times 3 p \times 3 q$
$(3 p + 3 q)^{2} = 9 p^{2} + 9 q^{2} + 18 p q$

Case 2:
1010 can be written as (1000 + 10) and 990 can be written as (1000 - 10)
$1010 \times 990 = (1000 + 10) \times (1000 - 10)$

Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ ,

$(1000 + 10) (1000 - 10)$ = $1000^{2}$ - $10^{2}$
= $1000000 - 100$
= $999900$

Case 3:
Using the identity, $(a - b)^{2} = a^{2} + b^{2} - 2 a b$

Here, $a = a^{2}$ and $b = b^{2}$
$(a^{2} - b^{2})^{2} = (a^{2})^{2} + (b^{2})^{2} - 2 a^{2} b^{2}$
$= a^{4} + b^{4} - 2 a^{2} b^{2}$

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