Simplify y^3 - 64 / y - 4

LMande

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Jul 6, 2006
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y^3 - 64 / y-4

My guess is that it's simplified? Am I right or wrong?
 
LMande said:
y^3 - 64 / y-4
The above means this:

. . . . .\(\displaystyle \L y^3 \,-\,\frac{64}{y} \,-\, 4\)

Is this what you meant? Or did you mean one of the following:

. . . . .\(\displaystyle \L y^3 \,-\, \frac{64}{y\,-\,4}\)

. . . . .\(\displaystyle \L \frac{y^3\,-\,64}{y\,-\,4}\)

...or something else?

Thank you.

Eliz.
 
please use proper grouping symbols ...
(y^3 - 64)/(y - 4)

y^3 - 64 = y^3 - 4^3, is the difference between two cubes ... which does factor.

fyi ... the factoring pattern is a^3 - b^3 = (a - b)(a^2 + ab + b^2)

also, the sum of two cubes is also factorable ...
a^3 + b^3 = (a + b)(a^2 - ab + b^2)

place these two algebra factoring facts into your algebra "toolbox" ... you will see them again.
 
Simplify...

Okay... so I'm REALLY lost!

Eliz... it is the last one that you listed that's my question. I'm going to attempt to write it again (with brackets) and I hope it makes sense.

(y^3 - 64) / (y-4)

I hope that makes more sense.
 
\(\displaystyle \L
y^3 - 64 = \left( {y - 4} \right)\left( {y^2 + 4y + 16} \right)\)
 
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