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Solve b^3=-8
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b=-2
-2 * -2 * -2 =-8
I suppose we had better go along with the idea that there are 3 roots for a cubic (b+2)(b^2-2b+4) gives in addition 2 complex roots i sqrt3 plus minus 1.
b^3 = (-2)^3 since it is an equality and the powers/indices r same => bases also have to be same therefore b = -2
b^3 - 8 = 0 is a suitable and permitted rearrangement that provides a polynomialk subject to the FTA.
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in junior classes, the concept of complex roots is not taught............. to me the questioner appeared to be from a junior class, so I advised accordingly....☺
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