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Question Number 109750 by mathdave last updated on 25/Aug/20

	Answered by 1549442205PVT last updated on 28/Aug/20
	$x^6 +27=(x^2 )^3 +3^3 =(x^2 +3)(x^4 −3x^2 +9) =(x^2 +3)([(x^2 +3)^2 −9x^2 ] =(x^2 +3)(x^2 +3x+3)(x^2 −3x+3) (1/(x^6 +27))=((ax+b)/(x^2 +3))+((cx+d)/(x^2 +3x+3))+((ex+f)/(x^2 −3x+3)) ⇔(1/(x^6 +27))=((ax^5 +bx^4 −3ax^3 −3bx^2 +9ax+9b)/(x^6 +27)) +((cx^5 +(d−3c)x^4 +(6c−3d)x^3 +(6d−9c)x^2 +(9c−9d)x+9d)/(x^6 +27)) +((ex^5 +(3e+f)x^4 +(3f+6e)x^3 +(9e+6f)x^2 +(9e+9f)x+9f)/(x^6 +27)) =(((a+c+e)x^5 +(b+d−3c+3e+f)x^4 +(−3a+6c−3d+6e+3f)x^3 +(−3b+9e+6f+6d−9c)x^2 +(9a+9c−9d+9e+9f)x+9b+9d+9f)/(x^6 +27)) ⇔ { ((a+c+e=0)),((b+d−3c=0)),((−3a+6c−3d+6e+3f=0)),((−3b−9c+6d+9e+6f=0)),((9a+9c−9d+9e+9f=0)),((9b+9d+9f=1)) :} (∗) We get the matrix of the system of six linear equations of degree 1 w.r.t six unknown a,b,c,d,e,f as [(1,0,1,0,1,0,0),(0,1,(−3),1,3,1,0),((−3),0,6,(−3),6,3,0),(0,(−3),(−9),6,9,6,0),(9,0,9,(−9),9,9,0),(0,9,0,9,0,9,1) ] Solve above system by using variable substitution, algebraic addition (or equivalent tranformation for matrixes) we get a=0,b=(1/(27)),c=(1/(18)),d=(4/(27)),f=((−2)/(27)),e=(1/(54)) (1/(x^6 +27))=(1/(27(x^2 +3)))+((x+2)/(54(x^2 +3x+3)))−((x−2)/(54(x^2 −3x+3))) Hence ((324)/(x^3 +27))=((12)/((x^2 +3)))+((6(x+2))/(x^2 +3x+3))−((6(x−2))/(x^2 −3x+3))$
	$x^{6} + 27 = {(x^{2})}^{3} + 3^{3} = (x^{2} + 3) (x^{4} - {3 x}^{2} + 9)$ $= (x^{2} + 3) ([{(x^{2} + 3)}^{2} - {9 x}^{2}]$ $= (x^{2} + 3) (x^{2} + 3 x + 3) (x^{2} - 3 x + 3)$ $\frac{1}{x^{6} + 27} = \frac{ax + b}{x^{2} + 3} + \frac{cx + d}{x^{2} + 3 x + 3} + \frac{ex + f}{x^{2} - 3 x + 3}$ $\Leftrightarrow \frac{1}{x^{6} + 27} = \frac{{ax}^{5} + {bx}^{4} - {3 ax}^{3} - {3 bx}^{2} + 9 ax + 9 b}{x^{6} + 27}$ $+ \frac{{cx}^{5} + (d - 3 c) x^{4} + (6 c - 3 d) x^{3} + (6 d - 9 c) x^{2} + (9 c - 9 d) x + 9 d}{x^{6} + 27}$ $+ \frac{{ex}^{5} + (3 e + f) x^{4} + (3 f + 6 e) x^{3} + (9 e + 6 f) x^{2} + (9 e + 9 f) x + 9 f}{x^{6} + 27}$ $= \frac{(a + c + e) x^{5} + (b + d - 3 c + 3 e + f) x^{4} + (- 3 a + 6 c - 3 d + 6 e + 3 f) x^{3} + (- 3 b + 9 e + 6 f + 6 d - 9 c) x^{2} + (9 a + 9 c - 9 d + 9 e + 9 f) x + 9 b + 9 d + 9 f}{x^{6} + 27}$ $\Leftrightarrow {\begin{cases} a + c + e = 0 \\ b + d - 3 c = 0 \\ - 3 a + 6 c - 3 d + 6 e + 3 f = 0 \\ - 3 b - 9 c + 6 d + 9 e + 6 f = 0 \\ 9 a + 9 c - 9 d + 9 e + 9 f = 0 \\ 9 b + 9 d + 9 f = 1 \end{cases} (*)$ $We get the matrix of the system of six$ $linear equations of degree 1 w . r . t six unknown$ $a, b, c, d, e, f as$ $[\begin{matrix} 1 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & - 3 & 1 & 3 & 1 & 0 \\ - 3 & 0 & 6 & - 3 & 6 & 3 & 0 \\ 0 & - 3 & - 9 & 6 & 9 & 6 & 0 \\ 9 & 0 & 9 & - 9 & 9 & 9 & 0 \\ 0 & 9 & 0 & 9 & 0 & 9 & 1 \end{matrix}]$ $Solve above system by using variable$ $substitution, algebraic addition (or equivalent$ $tranformation for matrixes) we get$ $a = 0, b = \frac{1}{27}, c = \frac{1}{18}, d = \frac{4}{27}, f = \frac{- 2}{27}, e = \frac{1}{54}$ $\frac{1}{x^{6} + 27} = \frac{1}{27 (x^{2} + 3)} + \frac{x + 2}{54 (x^{2} + 3 x + 3)} - \frac{x - 2}{54 (x^{2} - 3 x + 3)}$ $Hence$ $\frac{324}{x^{3} + 27} = \frac{12}{(x^{2} + 3)} + \frac{6 (x + 2)}{x^{2} + 3 x + 3} - \frac{6 (x - 2)}{x^{2} - 3 x + 3}$
	Commented by mathdave last updated on 25/Aug/20

	$t h a n k s t h i s w a s w o n d e r f u l l y d o n e b y a g r e a t m a n$ $. y o u h a v e t r i e d . u s h o u l d h a v e t h r o w m o r e l i g h t$ $o n h o w y o u c a m e a b o u t l e t t e r s f r o m d m a t r i x b o x$ $f o r p e o p l e t h a t a r e n o t p r o l i k e y o u . h w d i d y o u$ $g e t t h o s e l e t t e r s$
	Commented by 1549442205PVT last updated on 26/Aug/20

	$I used indefined - coefficient method$ $which stated that :$ $‘ ‘ two the polynomials are equal \forall x if$ $and only if the their corresponding$ $coefficients are equal^{″}$ $The system (*) can be solved by many$ $different ways : can use variable$ $substituting, algebraic plus, transform$ $ation equivalent matrix . . .$
	Answered by Her_Majesty last updated on 25/Aug/20

	$w r o n g$ $\frac{324}{x^{6} + 27} = \frac{324}{(x^{2} + 3) (x^{2} + 3 x + 3) (x^{2} - 3 x + 3)} =$ $= \frac{12}{x^{2} + 3} + \frac{6 (x + 2)}{x^{2} + 3 x + 3} - \frac{6 (x - 2)}{x^{2} - 3 x + 3}$ $n o w {i t}^{'} s e a s y t o s o l v e t h e i n t e g r a l b y u s i n g t h e$ $c o m m o n k n o w n f o r m u l a$
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