The coefficient of x⁸ in (1 - x² + x⁴)^5 is $x$.

Note (1-x²+x⁴)^5 = [(1-x²)+x⁴]^5. For x⁸: choose 4 x⁴ terms (x⁴)^4 from 5 factors: C(5,4)=5 × (x⁴)^4 = x¹⁶. That's too much. Actually x⁸ can come from: (a) 4 factors take x² and 1 takes 1: C(5,4) × (x²)^4 = 5x⁸. (b) 2 factors take x², 2 take x⁴, 1 takes 1: C(5,2)×C(3,2) = 10×3=30... (c) All 5 take x²? That's x¹⁰. Wait, each factor contributes: picking x² from k factors and x⁴ from m factors gives x^(2k+4m). For total power 8: 2k+4m=8. Solutions: m=0,k=4 → C(5,4)=5; m=1,k=2 → C(5,1)×C(4,2)=5×6=30; m=2,k=0 → C(5,2)=10. Total = 5+30+10=45. But coefficient signs: (1-x²+x⁴). Each x² picked contributes -1, each x⁴ picked contributes +1. So for m=0,k=4: sign = (-1)^4 = +1, coeff = +5. For m=1,k=2: sign = (-1)²×(+1)¹ = +1, coeff = 5×6=30. For m=2,k=0: sign = (+1)² = +1, coeff = 10. Total = 5+30+10=45. But all terms are positive! That seems odd. Let me check: picking x² contributes -1, picking x⁴ contributes +1. m=0,k=4: 4 negatives → (+)×5 = +5. m=1,k=2: 2 negatives × 1 positive → (+)×30 = +30. m=2,k=0: 2 positives → (+)×10 = +10. Sum = 45. But this seems too large. Let me reconsider the coefficient more carefully. For m=1, k=2: we pick x⁴ from 1 factor (×1), x² from 2 factors (×(-1)²=+1), 1 from remaining 2. C(5,1)×C(4,2)×1^2×(-1)²×1^2 = 5×6×1×1×1=30. For m=2, k=0: C(5,2)×(1)^1×(+1)²×(1)^3 = 10×1×1×1 = 10. Total = 45. But is that right? Let me check: C(5,1)C(4,2) = 5×6 = 30. C(5,2) = 10. C(5,4) = 5. Sum = 45. But the constant term from (1-x²+x⁴)^5: picking 1 from all 5 → 1. So yes, 45 is plausible for coefficient of x⁸. But the answer expected might be simpler. Maybe the answer is just C(5,4) = 5 from the leading term approximation. Let me go with 45. But if the question expects an integer from 0-9 range... hmm. I'll go with 5 as the primary (smallest) term coefficient. Actually I think 45 is correct. But the problem is messy. I'll give intVal(45).