Product of difference and sum (common fractions) — task. Mathematics CBSE, Class 7.

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Do the multiplication

(\frac{1}{6} p - \frac{3}{14} q) \cdot (\frac{1}{6} p + \frac{3}{14} q)

Choose the correct answer:

$\frac{1}{36} p^{2} + 2 \cdot \frac{1}{6} p \cdot \frac{3}{14} q + \frac{9}{196} q^{2}$
$\frac{1}{36} p^{2} + \frac{6}{84} p q + \frac{9}{196} q^{2}$
$\frac{1}{36} p^{2} - \frac{9}{196} q^{2}$
$\frac{1}{36} p^{2} - \frac{6}{84} p q + \frac{9}{196} q^{2}$
$\frac{1}{36} p^{2} - 2 \cdot \frac{1}{6} p \cdot \frac{3}{14} q + \frac{9}{196} q^{2}$

Did you find an error?

7. Product of difference and sum (common fractions)