SOLUTION: I need to factor h^3-25. I've tried but I always get h in there. I've done (h)(h+5)(h-5) or (h)(h^2-25) but I always get h^3-25h. There is no perfect cube so I am struggling. Thank

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Question 1204219: I need to factor h^3-25. I've tried but I always get h in there. I've done (h)(h+5)(h-5) or (h)(h^2-25) but I always get h^3-25h. There is no perfect cube so I am struggling. Thank you
Found 4 solutions by ikleyn, MathLover1, Alan3354, math_tutor2020:
Answer by ikleyn(52875)   (Show Source):
You can put this solution on YOUR website!
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I need to factor h^3-25. I've tried but I always get h in there. I've done (h)(h+5)(h-5) or (h)(h^2-25)
but I always get h^3-25h. There is no perfect cube so I am struggling. Thank you
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To factor   " h^3 - 25 "   is WRONG assignment.

It is NOT POSSIBLE to factor it with integer degrees and integer coefficients.

Even do not try.

Do not spend your time for nothing.

What is the source of your assignment ?

This source is FALSE, equally as this assignment itself.

Do not forget to post your THANKS to me for saving you from doing this non-sensical job.

Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!
...... rewrite as
.....now you have the difference of cubes and apply the rule
in your case and

Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
I need to factor h^3-25. I've tried but I always get h in there. I've done (h)(h+5)(h-5) or (h)(h^2-25) but I always get h^3-25h. There is no perfect cube so I am struggling. Thank you
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Maybe it's h^3 - 125?

Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

It's possible that the original expression is h^3 - 125

If so, then use the difference of cubes factoring rule
a^3 - b^3 = (a-b)(a^2 + ab + b^2)

to find that,
h^3 - 125 = (h-5)(h^2 + 5h + 25)