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How do you find the cube roots #root3(125)#?

Answer 1

Luna Angel

The answer is 5

#x# is a Real number and #n# is a Natural number bigger than 0 (#n=1, 2, 3, 4, ...#)

The rule is : #rootn(x^n)=x#

For example : #root2((5,3)^2)=5,3# or #root7((2/3)^7)=2/3#

We'll have instead : #rootn((-x)^n)=-rootn(x^n)=-x#

For example : #root3((-3)^3) = -3#

In this example, we need to write 125 in a form of #x^3#, since we have #n=3#, to do so we must think of writing it as a product of prime factors.

Five is the smallest prime number that divides by 125.

#125/5=25#

So #125=25xx5=5xx5xx5=5^3#

Now that you found the form #x^3#, apply the rule :

#root3(125)=root3(5^3)=5#

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Answer 2

Levi Adams

The cube root of 125, denoted as ( \sqrt[3]{125} ), can be found by calculating ( 125^{1/3} ).

[ \sqrt[3]{125} = 125^{1/3} ]

[ 125^{1/3} = 5 ]

So, the cube root of 125 is 5.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.