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Question
Express each of the following product as a monomials and verify the result for x = 1, y = 2:
\[\left( \frac{4}{9}ab c^3 \right) \times \left( - \frac{27}{5} a^3 b^2 \right) \times \left( - 8 b^3 c \right)\]
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Solution
To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., \[a^m \times a^n = a^{m + n}\]
We have:
\[\left( \frac{4}{9}ab c^3 \right) \times \left( - \frac{27}{5} a^3 b^2 \right) \times \left( - 8 b^3 c \right)\]
\[ = \left\{ \left( \frac{4}{9} \right) \times \left( - \frac{27}{5} \right) \times \left( - 8 \right) \right\} \times \left( a \times a^3 \right) \times \left( b \times b^2 \times b^3 \right) \times \left( c^3 \times c \right)\]
\[ = \left\{ \left( \frac{4}{9} \right) \times \left( - \frac{27}{5} \right) \times \left( - 8 \right) \right\} \times \left( a^{1 + 3} \right) \times \left( b^{1 + 2 + 3} \right) \times \left( c^{3 + 1} \right)\]
\[ = \frac{96}{5} a^4 b^6 c^4\]
Thus, the answer is \[\frac{96}{5} a^4 b^6 c^4\].
\[\because\] The expression doesn't consist of the variables x and y.
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