Express 1x3−1 in partial fractions
...Express 1x3−1 in partial fractions
13(1x−1−(x+2)x2+x+1)
13(1x−1−x−2x2+x+1)
13(1x−1−(x−2)x2+x+1)
13(1x−1−(x−1)x2−x−1)
Correct answer is A
1x3−1
x3−1=(x−1)(x2+x+1)
1x3−1=Ax−1+Bx+Cx2+x+1
1x3−1=A(x2+x+1)+(Bx+C)(x−1)x3−1
Comparing the two sides of the equation,
A+B=0...(1)
A−B+C=0...(2)
A−C=1...(3)
From (3), C=A−1, putting that in (2),
A−B=−C⟹A−B=1−A
2A−B=1...(4)
(1) + (4): 3A=1⟹A=13
A=−B⟹B=−13C=A−1⟹C=13−1=−23
= 13(1x−1−(x+2)x2+x+1)