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Which number in the monomial 125x^18y^3z^25 needs to be changed to make it a perfect cube

User Voltrevo
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2 Answers

2 votes

Hi,

You just have to change z^25 to z^24 or z^27.


125x^(18)y^3z^(24)=(5x^6*y*z^8)^3

User Geoff Norton
by
5.3k points
3 votes

Answer:

The required number in teh polynomial needs to be changes to make it a perfect cube is
z^(25) .

Explanation:

Given : Expression
125x^(18)y^3z^(25)

To find : Which number in the monomial expression needs to be changed to make it a perfect cube?

Solution :

Expression
125x^(18)y^3z^(25)

Using property,


(x^a)^b=x^(ab) and
(xy)^a=x^ay^a

Now we distribute each term into a power cube.


125=5^3


x^(18)=x^(3\cdot 6)


y^(3)=y^(3\cdot 1)


z^(25)=z^(3\cdot 8+1)

We have seen that


z^(25) is not making a multiple of 3 so to make it a perfect cube we have to change it into
z^(24)=Z^(3\cdot 8)

Now, Making a perfect cube with the change


125x^(18)y^3z^(25)


=5^3x^(3\cdot 6)y^(3\cdot 1)z^(3\cdot 8)


=(5)^3(x^(6))^3(y)^(3)(z^8)^3


=(5x^6yz^8)^3

Therefore, The required number in teh polynomial needs to be changes to make it a perfect cube is
z^(25) .

User Igloczek
by
4.9k points