Question

Find the product of the following monomials.
\[\begin{align}
  & \left( i \right)xy,{{x}^{2}}y,xy,x \\
 & \left( ii \right)a,b,ab,{{a}^{3}}b,a{{b}^{3}} \\
\end{align}\]

Answer 1

Hint: In this question, we are given monomials and we need to find their product. For this, we will first count the number of variables of each type from all monomials and then write the power of the variable as that number. For final answer, we will have product of the form ${{x}^{m}}\cdot {{y}^{n}}$ where m is number of x variable from all monomial and n is the number of y variables from all monomial.

Complete step by step answer:
Here we are given the monomials as,
1: $xy,{{x}^{2}}y,xy,x$.
Here, we have one x in the first monomial (xy).
We have two x in the second monomial $\left( {{x}^{2}}y \right)$.
We have one x in the third monomial (xy).
We have one x in the fourth monomial (x).
From all monomials, numbers of x are $1+2+1+1=5$.
So in the product, power of x will be 5. Similarly,
We have one y in the first monomial (xy).
We have one y in the second monomial $\left( {{x}^{2}}y \right)$.
We have one y in the third monomial (xy).
We have zero y in the fourth monomial (x).
From all monomials, numbers of y are $1+1+1+0=3$.
So in the product, power of y will be 3.
Therefore, our final monomial will look like this: ${{x}^{5}}\cdot {{y}^{3}}$.
Hence, $\left( xy \right)\left( {{x}^{2}}y \right)\left( xy \right)\left( x \right)={{x}^{5}}\cdot {{y}^{3}}$.
2: $a,b,ab,{{a}^{3}}b,a{{b}^{3}}$.
Here, we have one a in the first monomial (a).
We have zero a in the second monomial (b).
We have one a in the third monomial (ab).
We have three a in the fourth monomial $\left( {{a}^{3}}b \right)$.
We have one a in the fifth monomial $\left( a{{b}^{3}} \right)$.
From all monomials, numbers are $1+0+1+3+1=6$.
So in the product, power of a will be 6.
Similarly, we have zero b in the first monomial (a).
We have one b in the second monomial (b).
We have one b in the third monomial (ab).
We have one b in the fourth monomial $\left( {{a}^{3}}b \right)$.
We have three b in the fifth monomial $\left( a{{b}^{3}} \right)$.
From all monomials, numbers of b are $0+1+1+1+3=6$.
So in the product, power of b will be 6.
Therefore, our final monomial will look like this: ${{a}^{6}}\cdot {{b}^{6}}$.
Hence, $\left( a \right)\left( b \right)\left( ab \right)\left( {{a}^{3}}b \right)\left( a{{b}^{3}} \right)={{a}^{6}}\cdot {{b}^{6}}$.

Note: We can solve this question, using law of exponents ${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$ also. For (1), we have $xy,{{x}^{2}}y,xy,x$. It can be written as ${{x}^{1}}\cdot {{x}^{2}}\cdot {{x}^{1}}\cdot {{x}^{1}}\cdot {{y}^{1}}\cdot {{y}^{1}}\cdot {{y}^{1}}$. Using law of exponents, we have ${{x}^{1+2+1+1}}\cdot {{y}^{1+1+1}}={{x}^{5}}\cdot {{y}^{3}}$ which is the same answer.
Similarly, for (2), we have $a,b,ab,{{a}^{3}}b,a{{b}^{3}}$. It can be written as ${{a}^{1}}\cdot {{a}^{1}}\cdot {{a}^{3}}\cdot {{a}^{1}}\cdot {{b}^{1}}\cdot {{b}^{1}}\cdot {{b}^{1}}\cdot {{b}^{3}}={{a}^{1+1+3+1}}\cdot {{b}^{1+1+1+3}}={{a}^{6}}\cdot {{b}^{6}}$ which is the same answer.