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Solving Cubic Equations By Factoring: X^3+27=0
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i think its (x+3)(x^2-3x+9)=0
yea thats it, so now set x+3=0 and (x^2-3x+9)=0 and solve for x in both
-3 because x^3=-27 Take cube root of both side x= cube root of (-27)=-3
Brie, you would lose some of the solutions by doing that. this problem has more than 1 x
I have this: x+3=0 and (x^2-3x+9)=0 but how do you do it after this point?
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so for the left one you just subtract 3 and get x=-3....for the right one you would either use the quadratic equation or factor it first
Okay, Thank You!
np, but you would get imaginary numbers for the second one and I dont know if your teacher is looking for that or not
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