Question

How do you simplify ${\left( {4x{y^2}} \right)^3}$?

Answer 1

Hint: First of all, use the fact that ${\left( {abc} \right)^n} = {a^n}{b^n}{c^n}$ and then, just replace a by 4, b by x, c by ${y^2}$ and n by 3. Thus, the value of ${\left( {4x{y^2}} \right)^3}$ is obtained thereafter.

Complete step-by-step solution:
We are given that we are required to simplify ${\left( {4x{y^2}} \right)^3}$.
Since we know that we have a fact given by the following expression:-
$ \Rightarrow {\left( {abc} \right)^n} = {a^n}{b^n}{c^n}$
Replace a by 4, b by x and c by ${y^2}$ in the above mentioned expression, we will then obtain the following expression:-
$ \Rightarrow {\left( {4x{y^2}} \right)^n} = {4^n}{x^n}{\left( {{y^2}} \right)^n}$
Now, we will replace n by 3, we will then obtain the left hand side as the expression which we are required to simplify as follows:-
$ \Rightarrow {\left( {4x{y^2}} \right)^3} = {4^3}{x^3}{\left( {{y^2}} \right)^3}$
Simplifying the cube of 4 on the right hand side, we will then obtain the following expression as:-
$ \Rightarrow {\left( {4x{y^2}} \right)^3} = \left( {4 \times 4 \times 4} \right){x^3}{\left( {{y^2}} \right)^3}$
Calculating the required multiplication on the right hand side, we will then obtain the following expression as:-
$ \Rightarrow {\left( {4x{y^2}} \right)^3} = 64{x^3}{\left( {{y^2}} \right)^3}$
Simplifying the term with y on the right hand side, we will then obtain the following expression as-
$ \Rightarrow {\left( {4x{y^2}} \right)^3} = 64{x^3}{y^6}$
Thus, we have the required answer.

Note: The students must note that we have used an underlying fact in the last steps of the answer which states that: ${\left( {{x^a}} \right)^b} = {x^{ab}}$, here we just replace x by y, a by 2 and b by 3 and thus we have the step in the solution as we required.
The students must also commit to memory the following formula:-
$ \Rightarrow {\left( {abc} \right)^n} = {a^n}{b^n}{c^n}$
This formula is true because of associativity of real numbers, so it does not matter in which order we multiply them, therefore, the individual powers come in the picture here. Because when we multiply abc by other abc, we can exchange those to write a with a, b with b and so on. This is true for all a, b and c whether they are real numbers or complex numbers.