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Question

Break x4x3+1 into partial fractions.

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Solution

We know that,
a3+b3=(a+b)(a2ab+b2)
x3+1=(x+1)(x2x+1)
x4x3+1=x4(x+1)(x2x+1)
=x4[A(x+1)+Bx+1x2x+1]....(1)
1(x+1)(x2x+1)=A(x2x+1)+(Bx+C)(x+1)(x+1)(x2x+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(x2x+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=AB=1313=23
On putting value of A,B and C
x4x3+1=x4[13(x+1)+x+23(x2x+1)]

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