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First substitute: u = 1 - 3xNext calculate: du/dx = -3 .... therefore dx = (-1/3) * duNow re-arrange the expression: Integrate 1/u * ( -1/3)*duNext recall the integral of 1/x is the natural logarithm, and...

f(x) = 8x³ + 1 / x³ >>> f(x) = 8x³ + x^-3
f'(x) = 3(8x² ) + -3(x^-4) >>> f'(x) = 24...

f(x) = (x-3)(2x^2 - x - 1)f(x) = (x-3)(2x +1)(x-1)x = 3, -1/2, 1 are roots of the cubic equation.