$If a = 6 p and b = 3 p,$
$then find the expansion of$
$(a + b)^{2} + (a - b)^{2} .$

72 p^{2} - 18 q^{2}

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

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18 p^{2} + 72 q^{2}

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18 p^{2} - 72 q^{2}

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Solution

The correct option is B $72 p^{2} + 18 q^{2}$
Given,
$a = 6 p a n d b = 3 q$

We know that,
$(a + b)^{2} = a^{2} + 2 a b + b^{2} (a - b)^{2} = a^{2} - 2 a b + b^{2}$

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

On substituting the values of $a$ and $b$ , we get,
$(a + b)^{2} + (a - b)^{2} = (6 p + 3 q)^{2} + (6 p - 3 q)^{2} = 2 (6 p)^{2} + 2 (3 q)^{2} = 2 (36 p^{2}) + 2 (9 q)^{2} = 72 p^{2} + 18 q^{2}$

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a = 6 p

b = 3 p

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

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then find the expansion of

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