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factor W^3+8=
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= ?
That's the sum of two perfect cubes. Do you remember that formula?
w^3 + 8 = 0 w^3 = -8 third sqrt of -8 = -2
Yeah, its \[a^3 \pm b^3\]
I thought it was what went into 8
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The formula is: a3 + b3 = (a + b)(a2 – ab + b2)
\[(w \pm 3)(w^2 \pm w*3 + 9)\]
The second plus/minus should be upside down.
So it'd be \[\left( w+2 \right)\left( w ^{2}-2w+4 \right)\]
@Mooby: the b = 3, not 2.
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