Simplify each of the following.
(a) $ \left(\frac{1}{2} a-b\right)\left(\frac{1}{2} a+b\right)\left(\frac{1}{4} a^{2}+b^{2}\right) $
(b) $ \left(\frac{p}{2}-\frac{q}{3}\right)\left(\frac{p}{2}+\frac{q}{3}\right)\left(\frac{p^{2}}{4}+\frac{q^{2}}{9}\right)\left(\frac{p^{4}}{16}+\frac{a^{4}}{81}\right) $
(c) $ \left(a^{2}+1-a\right)\left(a^{2}-1+a\right) $
(d) $ \left(4 b^{2}+2 b+1\right)\left(4 b^{2}-2 b-1\right) $

To do:

We have to simplify the below expressions.

Solution:

We know that,

$(a+b)^2=a^2+2ab+b^2$

$(a-b)^2=a^2-2ab+b^2$

Therefore,

$ \begin{array}{l}
( a) \ \left(\frac{1}{2} a-b\right)\left(\frac{1}{2} a+b\right)\left(\frac{1}{4} a^{2} +b^{2}\right) =\left(\frac{1}{4} a^{2} -b^{2}\right)\left(\frac{1}{4} a^{2} +b^{2}\right)\\
=\left(\frac{1}{16} a^{4} -b^{4}\right)\\
( b) \ \left(\frac{p}{2} -\frac{q}{3}\right)\left(\frac{p}{2} +\frac{q}{3}\right)\left(\frac{p^{2}}{4} +\frac{q^{2}}{9}\right)\left(\frac{p^{4}}{16} +\frac{q^{4}}{81}\right) =\left(\frac{p^{2}}{4} -\frac{q^{2}}{9}\right)\left(\frac{p^{2}}{4} +\frac{q^{2}}{9}\right)\left(\frac{p^{4}}{16} +\frac{q^{4}}{81}\right)\\
=\left(\frac{p^{4}}{16} -\frac{q^{4}}{81}\right)\left(\frac{p^{4}}{16} +\frac{q^{4}}{81}\right)\\
=\left(\frac{p^{8}}{256} -\frac{q^{8}}{6561}\right)\\
( c) \ \left( a^{2} +1-a\right)\left( a^{2} -1+a\right) =\left[ a^{2} +( 1-a)\right]\left[ a^{2} -( 1-a)\right]\\
=a^{4} -( 1-a)^{2}\\
=a^{4} -\left( 1-2a+a^{2}\right)\\
=a^{4} -1+2a-a^{2}\\
=a^{4} -a^{2} +2a-1\\
( d) \ \left( 4b^{2} +2b+1\right)\left( 4b^{2} -2b-1\right) =\left[ 4b^{2} +( 2b+1)\right]\left[ 4b^{2} -( 2b+1)\right]\\
=\left( 4b^{2}\right)^{2} -( 2b+1)^{2}\\
=16b^{4} -\left[( 2b)^{2} +2\times 2b\times 1+1^{2}\right]\\
=16b^{4} -4b^{2} -4b-1
\end{array}$

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Updated on: 10-Oct-2022

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