Express \(\frac{1}{x^{3}-1}\) in partial fractions
\(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)
\(\frac{1}{3}(\frac{1}{x - 1} - \frac{x - 2}{x^{2} + x + 1})\)
\(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 2)}{x^{2} + x + 1})\)
\(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 1)}{x^{2} - x - 1})\)
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1/x^3-1
we all knw dat x^3-1=(x-1)(x^2+x+1)
1/x^3-1=1/(x-1)( x^2+x+1)
1/x^3-1=A/(x-1)+Bx+c/(x^2+x+1)
1=A (x^2+x+1)+(Bx+c) (x-1)
1=A (x^2+x+1);; when x=1
1=A(1^2+1+1);;;; 1=3A:; A=1/3
1=A (x^2+x+1)+(Bx+c) (x-1);;; when x=0
1=A(0+0+1)+(0+c)+(0-1)
1=A-C;; since A=1/3
1=1/3-c;; c=-2/3
expanding the equation
1=A (x^2+x+1)+(Bx+c) (x-1)
1=Ax^2+Ax+A+Bx^2-Bx+cx-c
therefore 0x^2=Ax^2+Bx^2;:: A+B=0
since A=1/3:; 1/3+B=0
B=-1/3
therefore A=1/3; B=-1/3;C=-2/3
1/x^3-1=A/(x-1)+Bx+c/(x^2+x+1)
insert d values of A,B and C in the equation to give u d ans, wich is
1/x^3-1=1/3(x-1) -(x+2)/3(x^2+x+1)
therefore d correct and is B
