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how to simplify (x^3-8)/x-2
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i felt this was ishaan94!!
\[x^3-8=(x-2)(x^2+2x+4)\] then cancel
x^3-8=(x-2)(x^2+2x+4) so cancel x-2
the top facotrs to (x-2)(x^2+2x+4) which means the (x-2) terms cancel out. Leaving only (x^2+2x+4)
difference of two cubes \[a^3-b^3=(a-b)(a^2+ab+b^2)\]
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i mean the (x-2) terms divide out
you get it?
the answer is x^2-6
no
no you can see that answer is x^2+2x+4 as x-2 cancels out
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i will write it out \[\frac{x^3-8}{x-2}=\frac{(x-2)(x^2+2x+4)}{x-2}=x^2+2x+4\]
@shana would you like to try it with numbers and see?
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