64 is the cube of 4 so:
x3 + 64 = (x + 4)(x2 - 4x + 16)
x3 + 4x2 - 16x - 64 = (x + 4)(x + 4)(x - 4).
(x - 8)(x2 + 8x + 64)
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...
3x3 + 192 3(x3 + 64) ========
x3 + 4x2 - 16x - 64 = (x + 4)(x + 4)(x - 4).
(x - 8)(x2 + 8x + 64)
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
x3 + 4x2 + x + 4 = (x + 4)(x2 + 1)
(x - 4)(x^2 + 4x + 16)
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
x3 + 4x2 + 16x + 64 =x2*(x + 4) + 16(x + 4) = (x + 4)*(x2 + 16) which has no further real factors.
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...
x^(5) + x^(2)= x^(2) ))x^(3) + 1) Think of '1' as '1^(3)' Hence x^(2)(x^(3) + 1^(3)) x^(2)(x + 1)(x^(2) - x + 1^(2)) or x^(2)(x + 1)(x^(2) - x + 1) Done!!!!!
x(x^2 + 2)