Prove that $(2 x - 3 y) (2 x + 3 y) = (2 x)^{2} - (3 y)^{2}$ geometrically.

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Solution

Step 1: Draw a square and cut into $3$ parts.
Step 2: There are $1$ hided square green and $2$ rectangles (pink, blue)
Step 3: Area of the full square $=$ $(2 x)^{2} - (3 y)^{2}$
Step 4: Now we have to find the area of rectangle as shown in the figure.
Step 5: Consider the area of pink rectangle $=$ length $\times$ breadth $=$ $2 x (2 x - 3 y)$
Step 6: Area of blue rectangle $=$ $3 y (2 x - 3 y)$
Step 7: Area of full square $=$ area of pink rectangle $+$ area of blue rectangle.
i.e., $(2 x)^{2} - (3 y)^{2} = 2 x (2 x - 3 y) + 3 y (2 x - 3 y)$
$(2 x)^{2} - (3 y)^{2} = (2 x + 3 y) (2 x - 3 y)$
Hence, geometrically we proved the identity $(2 x)^{2} - (3 y)^{2} = (2 x + 3 y) (2 x - 3 y)$ .

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Factorization:

$9 (2 x - 3 y)^{2} - 12 (2 x - 3 y) (2 x + 3 y) + 4 (2 x + 3 y)^{2}$

(2 x - 3 y)^{3} + (x + 3 y)^{3} + 3 (2 x - 3 y)^{2} (x + 3 y) + 3 (2 x - 3 y) (x + 3 y)^{2}

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