
How do you write the simplified form of \[-{{64}^{\dfrac{1}{3}}}\] ?
Answer
556.8k+ views
Hint: we can simplify this expression using basic exponent formulas. First we have to arrange the terms in such a way we can apply any exponent formulas i.e., we will rewrite the base to make it as an exponent and then we apply the formula needed. After that we have to solve accordingly to arrive at the solution.
Complete step-by-step solution:
Before going to solve let us know a few basic formulas of exponents and powers. They are
\[{{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}\]
\[{{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}\]
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]
\[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\]
\[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]
These are the basic formulas of exponents and we will use them in the problem where ever needed and also if the problem is not satisfying any formula we will make it into any one formula above to simplify it.
Given equation is
\[-{{64}^{ \dfrac{1}{3}}}\]
Now we will rewrite our base as power so we can make it as form \[{{\left( {{a}^{m}} \right)}^{n}}\]
After rewriting our base we will get
\[\Rightarrow -{{\left( {{\left( 4 \right)}^{3}} \right)}^{\dfrac{1}{3}}}\]
Now by using the formula we have for \[{{\left( {{a}^{m}} \right)}^{n}}\] we can rewrite our expression as
\[\Rightarrow -{{\left( 4 \right)}^{3\times \dfrac{1}{3}}}\]
By simplifying it we will get
\[\Rightarrow -{{\left( 4 \right)}^{1}}\]
\[\Rightarrow -4\]
So by simplifying the given expression we will get\[-4\] as the most simplified form.
Note: we can solve this by making our base as \[{{\left( 2 \right)}^{6}}\] and then applying the formula. In this method after rewriting the base we will get
\[\Rightarrow -{{\left( {{\left( 2 \right)}^{6}} \right)}^{\dfrac{1}{3}}}\]
After applying formula we will get the equation as
\[\Rightarrow -{{\left( 2 \right)}^{6\times \dfrac{1}{3}}}\]
By simplifying it we will get
\[\Rightarrow -{{\left( 2 \right)}^{2}}\]
As we know that \[{{2}^{2}}=4\] we can write the simplest form as
\[\Rightarrow -4\]
So we can do it in either way. But we have to be aware of exponents formulas.
Complete step-by-step solution:
Before going to solve let us know a few basic formulas of exponents and powers. They are
\[{{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}\]
\[{{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}\]
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]
\[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\]
\[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]
These are the basic formulas of exponents and we will use them in the problem where ever needed and also if the problem is not satisfying any formula we will make it into any one formula above to simplify it.
Given equation is
\[-{{64}^{ \dfrac{1}{3}}}\]
Now we will rewrite our base as power so we can make it as form \[{{\left( {{a}^{m}} \right)}^{n}}\]
After rewriting our base we will get
\[\Rightarrow -{{\left( {{\left( 4 \right)}^{3}} \right)}^{\dfrac{1}{3}}}\]
Now by using the formula we have for \[{{\left( {{a}^{m}} \right)}^{n}}\] we can rewrite our expression as
\[\Rightarrow -{{\left( 4 \right)}^{3\times \dfrac{1}{3}}}\]
By simplifying it we will get
\[\Rightarrow -{{\left( 4 \right)}^{1}}\]
\[\Rightarrow -4\]
So by simplifying the given expression we will get\[-4\] as the most simplified form.
Note: we can solve this by making our base as \[{{\left( 2 \right)}^{6}}\] and then applying the formula. In this method after rewriting the base we will get
\[\Rightarrow -{{\left( {{\left( 2 \right)}^{6}} \right)}^{\dfrac{1}{3}}}\]
After applying formula we will get the equation as
\[\Rightarrow -{{\left( 2 \right)}^{6\times \dfrac{1}{3}}}\]
By simplifying it we will get
\[\Rightarrow -{{\left( 2 \right)}^{2}}\]
As we know that \[{{2}^{2}}=4\] we can write the simplest form as
\[\Rightarrow -4\]
So we can do it in either way. But we have to be aware of exponents formulas.
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