25. \( \left(x^{\frac{1}{2}}+x^{\frac{1}{4}}-1\right)^{3} \).

To expand the expression \( \left(x^{\frac{1}{2}}+x^{\frac{1}{4}}-1\right)^{3} \), you can use the Binomial Theorem, which states that \( (a + b + c)^n \) can be expanded using combinations of the terms raised to various powers. For our expression, we can let \( a = x^{\frac{1}{2}} \), \( b = x^{\frac{1}{4}} \), and \( c = -1 \), and we'll expand it as: \[ (a + b + c)^3 = \sum_{i+j+k=3} \frac{3!}{i!j!k!} a^i b^j c^k \] Calculating this for all combinations \( i+j+k=3 \): 1. \( (3, 0, 0) \): \( (x^{\frac{1}{2}})^3 = x^{\frac{3}{2}} \) 2. \( (2, 1, 0) \): \( 3 \cdot (x^{\frac{1}{2}})^2 \cdot (x^{\frac{1}{4}}) = 3x^{1}x^{\frac{1}{4}} = 3x^{\frac{5}{4}} \) 3. \( (2, 0, 1) \): \( 3 \cdot (x^{\frac{1}{2}})^2 \cdot (-1) = -3x^{1} = -3x \) 4. \( (1, 2, 0) \): \( 3 \cdot (x^{\frac{1}{2}})(x^{\frac{1}{4}})^2 = 3x^{\frac{1}{2}}x^{\frac{1}{2}} = 3x^{1} \) 5. \( (1, 1, 1) \): \( 6 \cdot (x^{\frac{1}{2}})(x^{\frac{1}{4}})(-1) = -6x^{\frac{3}{4}} \) 6. \( (1, 0, 2) \): \( 3 \cdot (x^{\frac{1}{2}})(-1)^2 = 3x^{\frac{1}{2}} \) 7. \( (0, 3, 0) \): \( (x^{\frac{1}{4}})^3 = x^{\frac{3}{4}} \) 8. \( (0, 2, 1) \): \( 3 \cdot (x^{\frac{1}{4}})^2(-1) = -3x^{\frac{1}{2}} \) 9. \( (0, 1, 2) \): \( 3 \cdot (x^{\frac{1}{4}})(-1)^2 = 3x^{\frac{1}{4}} \) 10. \( (0, 0, 3) \): \( (-1)^3 = -1 \) Now we group all these terms, combining where necessary: - From \( (3,0,0) \): \( x^{\frac{3}{2}} \) - From \( (2,1,0) \) and \( (1,2,0) \): \( 3x^{\frac{5}{4}} - 3x^{1} + 3x^{1} = 3x^{\frac{5}{4}} \) - From \( (1,1,1) \): \( -6x^{\frac{3}{4}} \) - From \( (0,2,1) \): \( -3x^{\frac{1}{2}} \) - From \( (0,1,2) \): \( 3x^{\frac{1}{4}} \) - From \( (0,0,3) \): \( -1 \) So, combining all the results, we ultimately find: \[ \left(x^{\frac{1}{2}}+x^{\frac{1}{4}}-1\right)^{3} = x^{\frac{3}{2}} + 3x^{\frac{5}{4}} - 6x^{\frac{3}{4}} - 3x^{\frac{1}{2}} + 3x^{\frac{1}{4}} - 1 \] This is the expanded form of the given expression.