8 If \( x^{3}=a(x+1)^{3}+b(x+1)^{2}+c(x+1)+d \), find the values of \( a, b, c \) and \( d \)

To find the values of \( a, b, c, \) and \( d \) in the equation: \[ x^{3} = a(x+1)^{3} + b(x+1)^{2} + c(x+1) + d \] we can follow these steps: 1. **Expand the Right-Hand Side (RHS):** \[ \begin{align*} a(x+1)^3 &= a(x^3 + 3x^2 + 3x + 1) \\ b(x+1)^2 &= b(x^2 + 2x + 1) \\ c(x+1) &= c(x + 1) \\ \end{align*} \] Adding these together along with \( d \): \[ a(x^3 + 3x^2 + 3x + 1) + b(x^2 + 2x + 1) + c(x + 1) + d = a x^3 + (3a + b) x^2 + (3a + 2b + c) x + (a + b + c + d) \] 2. **Set the Expanded RHS Equal to the Left-Hand Side (LHS):** \[ x^3 = a x^3 + (3a + b) x^2 + (3a + 2b + c) x + (a + b + c + d) \] For the equation to hold for all \( x \), the coefficients of corresponding powers of \( x \) must be equal. This gives us the following system of equations: \[ \begin{cases} a = 1 \\ 3a + b = 0 \\ 3a + 2b + c = 0 \\ a + b + c + d = 0 \\ \end{cases} \] 3. **Solve the System of Equations:** - From the first equation: \( a = 1 \). - Substitute \( a = 1 \) into the second equation: \( 3(1) + b = 0 \) ⇒ \( b = -3 \). - Substitute \( a = 1 \) and \( b = -3 \) into the third equation: \( 3(1) + 2(-3) + c = 0 \) ⇒ \( 3 - 6 + c = 0 \) ⇒ \( c = 3 \). - Finally, substitute \( a = 1 \), \( b = -3 \), and \( c = 3 \) into the fourth equation: \( 1 - 3 + 3 + d = 0 \) ⇒ \( 1 + d = 0 \) ⇒ \( d = -1 \). **Conclusion:** \[ a = 1, \quad b = -3, \quad c = 3, \quad d = -1 \] **Verification:** Substituting \( a, b, c, \) and \( d \) back into the original equation confirms that both sides are equal. **Final Answer:** \[ \boxed{a=1,\quad b=-3,\quad c=3,\quad d=-1} \]