We know that, a3+b3=(a+b)(a2−ab+b2) x3+1=(x+1)(x2−x+1) x4x3+1=x4(x+1)(x2−x+1) =x4[A(x+1)+Bx+1x2−x+1]....(1) 1(x+1)(x2−x+1)=A(x2−x+1)+(Bx+C)(x+1)(x+1)(x2−x+1) 1=A(x2−(x+1)+(Bx+C)(x+1)....(2) Putting x=−1 in equation (2) (−1)4=A[(−1)2−(−1)+1]+(B(−1)+C)(−1+1) 1=A(1+1+1)+(−B+C)(0) 1=3A A=13 From equation (2) 1=A(x2−x+1)+Bx2+Bx+Cx+1 On comparing with coefficients of x2 0=A+B On putting value of A,0=13+B B=−13
Comparing coefficient of x, 0=−A+B+C
C=A−B=13−−13=23 On putting value of A,B and C x4x3+1=x4[13(x+1)+−x+23(x2−x+1)]