Exercise 2.77 in Nielsen and Chuang asks to show by example that there exist tripartite states $| \psi \rangle_{ABC} $ which cannot be written as
$$| \psi \rangle = \sum_i \lambda_i | i_A \rangle | i_B \rangle | i_C \rangle .$$
In this unofficial solution set, they consider the state
$$| \psi \rangle = | 0 \rangle_A \otimes | \Phi_+ \rangle_{BC} = |0 \rangle_A \otimes \bigg[ \frac{1}{\sqrt{2}} \Big( |00 \rangle_{BC} + | 11 \rangle_{BC} \Big) \bigg].$$
Then they perform the arbitrary (unitary) change of basis
$$|0\rangle = \alpha | \phi_0 \rangle + \beta | \phi_1 \rangle , \\ |1\rangle = \gamma | \phi_0 \rangle + \delta | \phi_1 \rangle $$
obtaining
$$ | \psi \rangle = \Big( \alpha | \phi_0 \rangle_A + \beta | \phi_1 \rangle_B \Big) \otimes \bigg[ \frac{1}{\sqrt{2}} \Big( |\phi_0 \phi_0 \rangle_{BC} + | \phi_1 \phi_1 \rangle_{BC} \Big) \bigg].$$
Since the cross terms cannot vanish, the proof is finished. However, I don't see the last step. Why do you have $ |00 \rangle_{BC} + | 11 \rangle_{BC} = |\phi_0 \phi_0 \rangle_{BC} + | \phi_1 \phi_1 \rangle_{BC}$ ? I mean, for instance, the unitarity of the change of basis doesn't buy us the cancellation of the terms $ | \phi_0 \phi_1 \rangle$ and $| \phi_1 \phi_0 \rangle$ (I think).
EDIT: Note that:
1) The coefficients $\alpha$, $\beta$, $\gamma$ and $\delta$ are not, in general, real. Since we want to show that there is no way of rewritting $| \psi \rangle$ in Schmidt form, we have to consider the most general case where these coefficients are complex.
2) The change of basis is, in the most general case, different for each party. I mean, we should rather write
$$|0 \rangle_X = \alpha_X \, | \phi_0 \rangle_X + \beta_X \, | \phi_1 \rangle_X \, , \\ | 1 \rangle_X = \gamma_X \, | \phi_0 \rangle_X + \delta_X \, | \phi_1 \rangle_X \, , $$
where $X= A, B, C$. Unitarity conditions are
$$| \alpha_X |^2 + |\beta_X |^2 = | \gamma_X |^2 + |\delta_X |^2 = 1 \, ,\\ \alpha_X^* \gamma_X + \beta_X^* \delta_X = 0 $$
for each $X$. I don't see how you can cancel the cross terms using this.