Question

How do you simplify \[\left( {5{x^3}} \right){\left( {2x} \right)^{ - 3}}\] ?

Answer 1

Hint: In this question, we are given an expression \[\left( {5{x^3}} \right){\left( {2x} \right)^{ - 3}}\] and we have to simplify it. First, we will multiply the powers inside the bracket, then we will rearrange the positions of the expressions to make it easier and then simplify it.

Formula used: \[\left( {{x^0}} \right) = 1\]
${x^m} \times {x^n} = {x^{m + n}}$

Complete step-by-step solution:
Here, we are given an expression \[\left( {5{x^3}} \right){\left( {2x} \right)^{ - 3}}\] and are asked to simplify it. Simplify is nothing but making the given expression simpler i.e., making it short. Now, we will do this by multiplying the two terms and making it as one term.
\[\left( {5{x^3}} \right){\left( {2x} \right)^{ - 3}}\]
First, we will multiply the powers inside the brackets,
= \[\left( {5{x^3}} \right)\left( {{2^{ - 3}}{x^{ - 3}}} \right)\]
Now, by applying the property of commutative property of multiplication, we get,
= \[\left( {5 \cdot {2^{ - 3}}} \right)\left( {{x^3}{x^{ - 3}}} \right)\]
We know that, if the given bases are the same, we can add the exponents, i.e., powers of the bases. We can see here, the bases are the same, i.e., $x$ so we will add its powers or exponentials.
= \[\left( {5 \cdot {2^{ - 3}}} \right)\left( {{x^{3 + ( - 3)}}} \right)\]
And it becomes,
=\[\left( {5 \cdot {2^{ - 3}}} \right)\left( {{x^{3 - 3}}} \right)\]
=\[\left( {5 \cdot {2^{ - 3}}} \right)\left( {{x^0}} \right)\]
We know that \[\left( {{x^0}} \right) = 1\] , so it becomes,
= \[\left( {5 \cdot {2^{ - 3}}} \right)\]
Now, there are two numbers in which one contains an exponential in negative, i.e., \[{2^{ - 3}}\] which means the cubic root of \[2\] . Since we can’t find the perfect cubic root for \[2\] , we can try to rewrite the term.
Multiplying and dividing by \[{2^3}\] , we get,
\[ = 5 \cdot {2^{ - 3}} \cdot \dfrac{{{2^3}}}{{{2^3}}}\]
= \[5 \cdot \dfrac{{{2^{ - 3}}{2^3}}}{{{2^3}}}\]
We know that, if the given bases are the same, we can add the exponents i.e., powers of the bases. We get,
= \[5 \cdot \dfrac{{{2^{ - 3}}^{ + 3}}}{{{2^3}}}\]
We know that anything to the power $0$ is $1$ , i.e., \[\left( {{x^0}} \right) = 1\] , where $x$ is any number, so it becomes
= \[5 \cdot \dfrac{{{2^0}}}{{{2^3}}}\]
= \[5 \cdot \dfrac{1}{{{2^3}}}\]
Now expanding \[{2^3}\] ,
= \[\dfrac{5}{{2 \times 2 \times 2}}\]
And now it becomes,
= \[\dfrac{5}{8}\]

Therefore, simplification of \[\left( {5{x^3}} \right){\left( {2x} \right)^{ - 3}}\] is \[\dfrac{5}{8}\].

Note: Alternative method:
The given term is \[\left( {5{x^3}} \right){\left( {2x} \right)^{ - 3}}\] , we need to simplify it.
First, we will multiply the powers inside the brackets,
= \[\left( {5{x^3}} \right)\left( {{2^{ - 3}}{x^{ - 3}}} \right)\]
Now, by applying the property of commutative property of multiplication we get,
= \[\left( {5 \cdot {2^{ - 3}}} \right)\left( {{x^3}{x^{ - 3}}} \right)\]
We will rewrite the expression by,
Multiplying and dividing by ${x^3}$ and \[{2^3}\] , we get
\[ = {5.2^{ - 3}}.\dfrac{{{2^3}}}{{{2^3}}}{\text{ }} \times {\text{ }}{x^3}{x^{ - 3}}\dfrac{{{x^3}}}{{{x^3}}}\]
= \[5 \cdot \dfrac{{{2^{ - 3}}{2^3}}}{{{2^3}}} \times {x^3}\dfrac{{{x^{ - 3}}{x^3}}}{{{x^3}}}\]
We know that, if the given bases are the same, we can add the exponents i.e., powers of the bases. We get,
= \[5 \cdot \dfrac{{{2^{ - 3}}^{ + 3}}}{{{2^3}}}{x^3}\dfrac{{{x^{ - 3 + }}^3}}{{{x^3}}}\]
We know that anything to the power $0$ is $1$ , i.e., \[\left( {{x^0}} \right) = 1\] , where $x$ is any number, so it becomes
= \[5 \cdot \dfrac{{{2^0}}}{{{2^3}}}{x^3}\dfrac{{{x^0}}}{{{x^3}}}\]
= \[5 \cdot \dfrac{1}{{{2^3}}}{x^3}\dfrac{1}{{{x^3}}}\]
= \[\dfrac{5}{{{2^3}}}.\dfrac{{{x^3}}}{{{x^3}}}\]
Now expanding \[{2^3}\] ,
= \[\dfrac{5}{{2 \times 2 \times 2}}\]
And now it becomes,
= \[\dfrac{5}{8}\]
Therefore, simplification of \[\left( {5{x^3}} \right){\left( {2x} \right)^{ - 3}}\] is \[\dfrac{5}{8}\] .