Question 79976
Can someone show me the factoring steps that gets h^3+8/h+2 to h^2-2h+4? 
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Somewhere in your algebra journey, you may have come across a "Special factoring Formula" section, (a good thing to know).
One of these formulas is called the "Sum of Cubes"
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It is often given as: u^3 + v^3 = (u + v)(u^2 - uv + v^2)
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h^3 + 8 is the sum of cubes but in this case u = h and v = 2; so we have:
{{{((h + 2)(h^2 - 2h + 4))/((h + 2))}}}
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I'm sure you can see what happens now.