Express $\frac{1}{9} x^{2} - \frac{1}{16} y^{2}$ as a product of binomials and find the value if x = 3 and y = 4.

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Solution

The correct option is A
0

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

The algebraic expression is $\frac{1}{9} x^{2} - \frac{1}{16} y^{2}$ .

$= (\frac{1}{3} x)^{2} - (\frac{1}{4} y)^{2} = (\frac{1}{3} x + \frac{1}{4} y) (\frac{1}{3} x - \frac{1}{4} y)$

Put x = 3 and y = 4.

$= (\frac{3}{3} + \frac{4}{4}) (\frac{3}{3} - \frac{4}{4})$

$= (1 + 1) (1 - 1) = 2 \times 0 = 0$

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