Given, $$\left| {\matrix{ x & {{x^2}} & {1 + {x^3}} \cr {2x} & {4{x^2}} & {1 + 8{x^3}} \cr {3x} & {9{x^2}} & {1 + 27{x^3}} \cr } } \right| = 10$$

$$ \Rightarrow x\,.\,{x^2}\left| {\matrix{ 1 & 1 & {1 + {x^3}} \cr 2 & 4 & {1 + 8{x^3}} \cr 3 & 9 & {1 + 27{x^3}} \cr } } \right| = 10$$

Apply R₂ $$\to$$ R₂ $$-$$ 2R₁ and R₃ $$\to$$ R₃ $$-$$ 3R₁, we get

$${x^3}\left| {\matrix{ 1 & 1 & {1 + {x^3}} \cr 0 & 2 & { - 1 + 6{x^3}} \cr 0 & 6 & { - 2 + 24{x^3}} \cr } } \right| = 10$$

$$ \Rightarrow {x^3}\,.\,\left| {\matrix{ 2 & {6{x^2} - 1} \cr 6 & {24{x^3} - 2} \cr } } \right| = 10$$

$$ \Rightarrow {x^3}(48{x^3} - 4 - 36{x^3} + 6) = 10$$

$$ \Rightarrow 12{x^6} + 2{x^3} = 10$$

$$ \Rightarrow 6{x^6} + {x^3} - 5 = 0$$

$$ \Rightarrow 6{({x^3})^2} + {x^3} - 5 = 6$$

$$ \Rightarrow 6{({x^3})^2} + 6{x^3} - 5{x^3} - 5 = 0$$

$$ \Rightarrow 6{x^3}({x^3} + 1) - 5({x^3} + 1) = 0$$

$$ \Rightarrow (6{x^3} - 5)({x^2} - x + 1)(x + 1) = 0$$

$$\therefore$$ $$x = {\left( {{5 \over 6}} \right)^{1/3}}, - 1$$

Hence, the number of real solutions is 2.