64 is the cube of 4 so:
x3 + 64 = (x + 4)(x2 - 4x + 16)
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(x - 8)(x2 + 8x + 64)
x3 + 4x2 - 16x - 64 = (x + 4)(x + 4)(x - 4).
x3+27y3 = (x+3y) · (x2-3xy+9y2)
x3 + 12x2 + 48x + 64= (x + 4)(x2 + 8x + 16)= (x + 4)(x + 4)(x + 4)= (x + 4)3
Subtract 1 from each side: x3 = 64; You might know that 4 cubed is 64, but if you don't then take log 64, divide it by 3 and take the antilog of your answer...