Answer
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Hint: We try to form the indices formula for the value 3. We first convert the decimal form into fraction form. Then we find the cube roots separately. This is a cube root of 15625. We find the prime factorisation of 15625. Then we take one digit out of the three same number of primes. There will be no odd number of primes remaining in the root which can’t be taken out.
Complete step-by-step answer:
We first convert the decimal form into fraction form and get $ -15.625=-\dfrac{15625}{1000} $ .
We need to find $ \sqrt[3]{-\dfrac{15625}{1000}}=-\sqrt[3]{\dfrac{15625}{1000}} $ . We know that $ {{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}} $ .
Therefore, $ -\sqrt[3]{\dfrac{15625}{1000}}=-\dfrac{\sqrt[3]{15625}}{\sqrt[3]{1000}} $ .
We also have that $ \sqrt[3]{1000}=\sqrt[3]{10\times 10\times 10}=10 $ .
We now find the cube root of 15625.
We need to find the value of the algebraic form of $ \sqrt[3]{15625} $ . This is a cube root form.
The given value is the form of indices. We are trying to find the cube root value of 15625.
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 factorisation of the given number 15625.
$ \begin{align}
& 5\left| \!{\underline {\,
15625 \,}} \right. \\
& 5\left| \!{\underline {\,
3125 \,}} \right. \\
& 5\left| \!{\underline {\,
625 \,}} \right. \\
& 5\left| \!{\underline {\,
125 \,}} \right. \\
& 5\left| \!{\underline {\,
25 \,}} \right. \\
& 5\left| \!{\underline {\,
5 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align} $
Therefore, \[15625=5\times 5\times 5\times 5\times 5\times 5\].
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 \[15625=5\times 5\times 5\times 5\times 5\times 5\], we will take out two 3 from the multiplication.
So, $ \sqrt[3]{15625}=\sqrt[3]{5\times 5\times 5\times 5\times 5\times 5}=5\times 5=25 $ .
So, the cube root of $ -15.625 $ is $ -\dfrac{25}{10}=-2.5 $ .
So, the correct answer is “-2.5”.
Note: We can also use the variable form where we can take $ x=\sqrt[3]{27} $ . But we need to remember that we can’t use the cube on both sides of the equation $ x=\sqrt[3]{27} $ 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 first convert the decimal form into fraction form and get $ -15.625=-\dfrac{15625}{1000} $ .
We need to find $ \sqrt[3]{-\dfrac{15625}{1000}}=-\sqrt[3]{\dfrac{15625}{1000}} $ . We know that $ {{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}} $ .
Therefore, $ -\sqrt[3]{\dfrac{15625}{1000}}=-\dfrac{\sqrt[3]{15625}}{\sqrt[3]{1000}} $ .
We also have that $ \sqrt[3]{1000}=\sqrt[3]{10\times 10\times 10}=10 $ .
We now find the cube root of 15625.
We need to find the value of the algebraic form of $ \sqrt[3]{15625} $ . This is a cube root form.
The given value is the form of indices. We are trying to find the cube root value of 15625.
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 factorisation of the given number 15625.
$ \begin{align}
& 5\left| \!{\underline {\,
15625 \,}} \right. \\
& 5\left| \!{\underline {\,
3125 \,}} \right. \\
& 5\left| \!{\underline {\,
625 \,}} \right. \\
& 5\left| \!{\underline {\,
125 \,}} \right. \\
& 5\left| \!{\underline {\,
25 \,}} \right. \\
& 5\left| \!{\underline {\,
5 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align} $
Therefore, \[15625=5\times 5\times 5\times 5\times 5\times 5\].
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 \[15625=5\times 5\times 5\times 5\times 5\times 5\], we will take out two 3 from the multiplication.
So, $ \sqrt[3]{15625}=\sqrt[3]{5\times 5\times 5\times 5\times 5\times 5}=5\times 5=25 $ .
So, the cube root of $ -15.625 $ is $ -\dfrac{25}{10}=-2.5 $ .
So, the correct answer is “-2.5”.
Note: We can also use the variable form where we can take $ x=\sqrt[3]{27} $ . But we need to remember that we can’t use the cube on both sides of the equation $ x=\sqrt[3]{27} $ 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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