Multiply:
$ \left(\frac{-a}{7}+\frac{a^{2}}{9}\right) $ by $ \left(\frac{b}{2}-\frac{b^{2}}{3}\right) $

To do:

We have to multiply \( \left(\frac{-a}{7}+\frac{a^{2}}{9}\right) \) by \( \left(\frac{b}{2}-\frac{b^{2}}{3}\right) \)

Solution:

We know that,

$(a+b)\times(c+d)=a(c+d)+b(c+d)$

Therefore,

$(\frac{-a}{7}+\frac{a^{2}}{9}) \times(\frac{b}{2}-\frac{b^{2}}{3})=\frac{-a}{7} \times(\frac{b}{2}-\frac{b^{2}}{3})+\frac{a^{2}}{9} \times(\frac{b}{2}-\frac{b^{2}}{3})$

$=\frac{-a}{7} \times \frac{b}{2}+\frac{a}{7} \times \frac{b^{2}}{3}+\frac{a^{2}}{9} \times \frac{b}{2}-\frac{a^{2}}{9} \times\frac{b^{2}}{3}$

$=\frac{-a b}{14}+\frac{a b^{2}}{21}+\frac{a^{2} b}{18}-\frac{a^{2} b^{2}}{27}$

$=\frac{-1}{14} a b+\frac{1}{21} a b^{2}+\frac{1}{18} a^{2} b-\frac{1}{27} a^{2} b^{2}$

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Updated on: 10-Oct-2022

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