
What is the cube root of \[27{a^{12}}\]?
Answer
512.4k+ views
Hint: Here in this question, we have to find the cube root for the given question. Here the question includes algebraic expression, here the cube root has to be determined for the variable term also. To solve this we have to use the simple arithmetic operations and hence we determine the solution for the given question.
Complete step-by-step solutions:
In mathematics, a square root of a number x is a number y such that \[{y^2} = x\]; in other words, a number y whose square is x. Generally, the square root is denoted as \[\sqrt {} \]
In mathematics, a cube root of a number x is a number y such that $y^3 = x$. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. Generally, the cube root is denoted as \[\sqrt[3]{{}}\]
Now consider the given algebraic expression
\[ \Rightarrow 27{a^{12}}\]
The number 27 and variable \[{a^{12}}\] can be written as
\[ \Rightarrow \left( {3 \times 3 \times 3} \right)\left( {{a^4} \times {a^4} \times {a^4}} \right)\]
The above term can be written in the form of exponents and so we have
\[ \Rightarrow {3^3}{\left( {{a^4}} \right)^3}\]
The power term is written based on how many times the number is multiplied.
Here in this question we have to find the cube root, now we will apply the cube root for the above expression
\[ \Rightarrow \sqrt[3]{{{3^3}{{\left( {{a^4}} \right)}^3}}}\]
If the power is 3 then we call it as cube
Now we will take the cube root for each term
\[ \Rightarrow \sqrt[3]{{{3^3}}}\sqrt[3]{{{{\left( {{a^4}} \right)}^3}}}\]
The cube root and the cube are inverse to each other and it will gets cancels and so we have
\[ \Rightarrow 3\left( {{a^4}} \right)\]
Therefore the cube root of \[27{a^{12}}\] is \[3\left( {{a^4}} \right)\].
Note: When we want to find the cube root of some number, let it be x. If x is a perfect cube then we can obtain the result directly. Otherwise if x is not a perfect cube let us factorise the x and if it possible we write the number in the form of exponential and then we simplify the number.
Complete step-by-step solutions:
In mathematics, a square root of a number x is a number y such that \[{y^2} = x\]; in other words, a number y whose square is x. Generally, the square root is denoted as \[\sqrt {} \]
In mathematics, a cube root of a number x is a number y such that $y^3 = x$. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. Generally, the cube root is denoted as \[\sqrt[3]{{}}\]
Now consider the given algebraic expression
\[ \Rightarrow 27{a^{12}}\]
The number 27 and variable \[{a^{12}}\] can be written as
\[ \Rightarrow \left( {3 \times 3 \times 3} \right)\left( {{a^4} \times {a^4} \times {a^4}} \right)\]
The above term can be written in the form of exponents and so we have
\[ \Rightarrow {3^3}{\left( {{a^4}} \right)^3}\]
The power term is written based on how many times the number is multiplied.
Here in this question we have to find the cube root, now we will apply the cube root for the above expression
\[ \Rightarrow \sqrt[3]{{{3^3}{{\left( {{a^4}} \right)}^3}}}\]
If the power is 3 then we call it as cube
Now we will take the cube root for each term
\[ \Rightarrow \sqrt[3]{{{3^3}}}\sqrt[3]{{{{\left( {{a^4}} \right)}^3}}}\]
The cube root and the cube are inverse to each other and it will gets cancels and so we have
\[ \Rightarrow 3\left( {{a^4}} \right)\]
Therefore the cube root of \[27{a^{12}}\] is \[3\left( {{a^4}} \right)\].
Note: When we want to find the cube root of some number, let it be x. If x is a perfect cube then we can obtain the result directly. Otherwise if x is not a perfect cube let us factorise the x and if it possible we write the number in the form of exponential and then we simplify the number.
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