Show that $(2 a + 3 b)^{2} - 24 a b = (2 a - 3 b)^{2}$

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Solution

$(2 a + 3 b)^{2} - 24 a b = (2 a - 3 b)^{2}$
LHS
$= (2 a + 3 b)^{2} - 24 a b$
$\Rightarrow (2 a)^{2} + (3 b)^{2} + 2 (2) (3) (a) (b) - 24 a b$
$\Rightarrow 4 a^{2} + 9 b^{2} + 12 a b - 24 a b$
$\Rightarrow (2 a)^{2} + 4 (3 b)^{2} - 2 (2 a) (3 b)$
$\Rightarrow (2 a - 3 b)^{2}$ = RHS
$∴ (2 a + 3 b)^{2} - 24 a b = (2 a - 3 b)^{2}$

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Similar questions

What is the LCM of $(2 a + 3 b)^{2}$ and $(4 a^{2} - 9 b^{2})$

2 a + 6 b - 3 (a + 3 b)^{2}

2 a + 6 b - 3 (a + 3 b)^{2}

= 2 (a + 3 b) - 3 (a + 3 b)^{2} [T a k i n g 2 c o m m o n f r o m 2 a + 6 b]

= (a + 3 b) {\begin{matrix} 2 - 3 (a + 3 b) \end{matrix}} [T a k i n g (a + 3 b) c o m m o n]

= (a + 3 b) (2 - 3 a - 9 b)

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

Factorise:

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

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

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