How do you simplify \[\sqrt[3]{432}\]?
Answer
584.4k+ views
Hint: We try to form the indices formula for the value 3. This is a cube root of 432. We find the prime factorization of 432. Then we take one digit out of the three same number of primes. There will be some odd number of primes remaining in the root which can’t be taken out. We keep them as it is.
Complete step by step answer:
We need to find the value of the algebraic form of \[\sqrt[3]{432}\]. This is a cube root form.
The given value is the form of indices. We are trying to find the root value of 432.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 3 we get \[{{a}^{\dfrac{1}{3}}}=\sqrt[3]{a}\].
We need to find the prime factorization of the given number 432.
$ \begin{align}
& 2\left| \!{\underline {\,
432 \,}} \right. \\
& 2\left| \!{\underline {\,
216 \,}} \right. \\
& 2\left| \!{\underline {\,
108 \,}} \right. \\
& 2\left| \!{\underline {\,
54 \,}} \right. \\
& 3\left| \!{\underline {\,
27 \,}} \right. \\
& 3\left| \!{\underline {\,
9 \,}} \right. \\
& 3\left| \!{\underline {\,
3 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align} $
Therefore, \[432=2\times 2\times 2\times 2\times 3\times 3\times 3\].
For finding the cube root, we need to take one digit out of the three same number of primes.
This means in the cube root value of \[432=2\times 2\times 2\times 2\times 3\times 3\times 3\], we will take out one 2 and one 3 but one 2 isn’t in triplet form. In that case we keep the number inside the root.
So, $ \sqrt[3]{432}=\sqrt[3]{2\times 2\times 2\times 2\times 3\times 3\times 3}=2\times 3\times \sqrt[3]{2}=6\sqrt[3]{2} $ . Basically 432 is the cube of $ 6\sqrt[3]{2} $ .
We can also use the theorem of indices \[{{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\]. We know that $ 432={{\left( 6\sqrt[3]{2} \right)}^{3}} $ .
We need to find \[\sqrt[3]{432}\] which gives $ 432={{\left( 6\sqrt[3]{2} \right)}^{3}}={{\left[ {{\left( 6\sqrt[3]{2} \right)}^{3}} \right]}^{\dfrac{1}{3}}}={{\left( 6\sqrt[3]{2} \right)}^{3\times \dfrac{1}{3}}}=6\sqrt[3]{2} $ .
Therefore, the value of \[\sqrt[3]{432}\] is $ 6\sqrt[3]{2} $ .
Note:
We can also use the variable form where we can take $ x=\sqrt[3]{432} $ . But we need to remember that we can’t use the cube on both sides of the equation $ x=\sqrt[3]{432} $ as in that case we are taking two extra values as a root value. Then this linear equation becomes a cubic equation.
Complete step by step answer:
We need to find the value of the algebraic form of \[\sqrt[3]{432}\]. This is a cube root form.
The given value is the form of indices. We are trying to find the root value of 432.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 3 we get \[{{a}^{\dfrac{1}{3}}}=\sqrt[3]{a}\].
We need to find the prime factorization of the given number 432.
$ \begin{align}
& 2\left| \!{\underline {\,
432 \,}} \right. \\
& 2\left| \!{\underline {\,
216 \,}} \right. \\
& 2\left| \!{\underline {\,
108 \,}} \right. \\
& 2\left| \!{\underline {\,
54 \,}} \right. \\
& 3\left| \!{\underline {\,
27 \,}} \right. \\
& 3\left| \!{\underline {\,
9 \,}} \right. \\
& 3\left| \!{\underline {\,
3 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align} $
Therefore, \[432=2\times 2\times 2\times 2\times 3\times 3\times 3\].
For finding the cube root, we need to take one digit out of the three same number of primes.
This means in the cube root value of \[432=2\times 2\times 2\times 2\times 3\times 3\times 3\], we will take out one 2 and one 3 but one 2 isn’t in triplet form. In that case we keep the number inside the root.
So, $ \sqrt[3]{432}=\sqrt[3]{2\times 2\times 2\times 2\times 3\times 3\times 3}=2\times 3\times \sqrt[3]{2}=6\sqrt[3]{2} $ . Basically 432 is the cube of $ 6\sqrt[3]{2} $ .
We can also use the theorem of indices \[{{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\]. We know that $ 432={{\left( 6\sqrt[3]{2} \right)}^{3}} $ .
We need to find \[\sqrt[3]{432}\] which gives $ 432={{\left( 6\sqrt[3]{2} \right)}^{3}}={{\left[ {{\left( 6\sqrt[3]{2} \right)}^{3}} \right]}^{\dfrac{1}{3}}}={{\left( 6\sqrt[3]{2} \right)}^{3\times \dfrac{1}{3}}}=6\sqrt[3]{2} $ .
Therefore, the value of \[\sqrt[3]{432}\] is $ 6\sqrt[3]{2} $ .
Note:
We can also use the variable form where we can take $ x=\sqrt[3]{432} $ . But we need to remember that we can’t use the cube on both sides of the equation $ x=\sqrt[3]{432} $ as in that case we are taking two extra values as a root value. Then this linear equation becomes a cubic equation.
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