Let        g(x) = $\int$ p(x)dx

                    = x(x⁸ – 3x⁵ – 5x³ + 15) + c

                    = x(x³ – 3) (x⁵ – 5) + c

Clearly  f(0) = f(3^1/3) = f(5^1/5) = c

$\therefore$        Rolle's theorem is applicable on g(x) in intervals [0, 3^1/3], [3^1/3,5^1/5] and in [0, 5^1/5]

$\because$        y = x^1/x has a local maxima at x = e

$\therefore$        0 < 5^1/5 < 3^1/3

$\therefore$        There exists at least one 'c', each in [0, 5^1/5] and in [5^1/5, 3^1/3] for which g'(x) = 0 i.e.  p(x) = 0

$\therefore$        p(x) = 0 has at least two real roots in [0, 3^1/3].