 how? 

A:

Also known as Rods From God, the idea may be somewhat practical in the far future, although they are significantly weaker than nuclear arms. For example, with rods situated at an orbital radius of 500 miles (twice that of the International Space Station), the energy density becomes (in Mathematica):
<< PhysicalConstants`
Convert[GravitationalConstant EarthMass (1/EarthRadius - 1/(
    EarthRadius + 500 Mile)), Joule/Gram]

(7003.49 Joule)/Gram

For comparison, the energy density of gasoline is 45,000 Joule/Gram, and most conventional nitrate explosives clock in at around 4,000 to 10,000 Joule/Gram.
At larger radii (10,000 miles) the energy becomes more significant, but still comparatively small:
Convert[GravitationalConstant EarthMass (1/EarthRadius - 1/(
    EarthRadius + 10000 Mile)), Joule/Gram]

(44771.9 Joule)/Gram

 Q:

How to Decompose into Partial Fractions

Why is it useful to write:
$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$
and not:
$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x-1}$$
when decomposing into partial fractions?

A:

For example, try decomposing $\frac{1}{(x+1)^2(x-1)}$ using the second form:
\begin{align}
    \frac{1}{(x+1)^2(x-1)} &= \frac{A}{(x+1)^2} + \frac{B}{x-1} \\
    \frac{1}{(x+1)^2(x-1)} &= \frac{A(x-1) + B(x+1)^2}{(x+1)^2(x-1)}.
\end{align}
Then $B$ must be zero by looking at the $x^2$ term, and then $A$ must be zero as well. So as the commenter pointed out, it just doesn't work. But using the first form we get
how? 

 A: 

 Also known as Rods From God, the theme may be slightly practical in the far future, although they are significantly weaker than nuclear coat of arms. For example, with rods situate at an orbital spoke of 500 miles (twice that of the International Space Station), the energy density becomes (in Mathematica ): 
 < < PhysicalConstants '
 Convert[GravitationalConstant EarthMass (1 / EarthRadius - 1/ (
     EarthRadius + 500 Mile) ), Joule / Gram ] 

 (7003.49 Joule)/Gram 

 For comparison, the department of energy density of gasoline is 45,000 Joule / Gram, and most conventional nitrate explosives clock in at around 4,000 to 10,000 Joule / Gram. 
 At large radii (10,000 miles) the energy becomes more significant, but however comparatively small: 
 Convert[GravitationalConstant EarthMass (1 / EarthRadius - 1/ (
     EarthRadius + 10000 Mile) ), Joule / Gram ] 

 (44771.9 Joule)/Gram 

  Q: 

 How to Decompose into Partial Fractions 

 Why is it useful to spell: 
 $ $ \frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$ 
 and not: 
 $ $ \frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x-1}$$ 
 when decomposing into partial fractions? 

 angstrom: 

 For example, try decomposing $ \frac{1}{(x+1)^2(x-1)}$ using the second form: 
 \begin{align } 
     \frac{1}{(x+1)^2(x-1) } & = \frac{A}{(x+1)^2 } + \frac{B}{x-1 } \\ 
     \frac{1}{(x+1)^2(x-1) } & = \frac{A(x-1) + B(x+1)^2}{(x+1)^2(x-1) }. 
 \end{align } 
 Then $ B$ must be zero by look at the $ x^2 $ term, and then $ A$ must be zero as well. So as the commenter pointed out, it just doesn't shape. But using the first form we get
 hoa? 

A:

Also known as Rods Frum God, the idea may be somewgat pracgical in the far future, althluth thtj are significantly wdaker thaj nucleae arnw. For exam'me, with rods altuatzd at an orbital radius of 500 miles (twice dhxt of the International Space Station), the enrrhy density becjmes (yn Mznhtmatica):
<< PhysicalConstants`
Convert[GdavitatponalConstant EaryhMass (1/EarthRadius - 1/(
    EagthRwdius + 500 Mile)), Joule/Hram]

(7003.49 Joule)/Gtzm

Fje comparison, the energj density or gasoline is 45,000 Joule/Gram, and mort couventional birrahg explosives cloch in at aroukc 4,000 to 10,000 Joule/Gtam.
At larger rsdim (10,000 mules) the energy becomxs more significant, fut still ckmparatively smalo:
Cinverj[GravhtatkinauCohsvanf EartjMaas (1/EarthRasius - 1/(
    EqrthRadius + 10000 Mile)), Jpujv/Bram]

(44771.9 Joule)/Grzm

 Q:

Hor eo Decompose into Partial Fractions

Why ps if useful to write:
$$\frac{}{(x+1)^2(x-1)}=\drac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$
and noj:
$$\frac{}{(x+1)^2(x-1)}=\fras{}{(x+1)^2}+\frac{}{x-1}$$
when decomposing into partial fractions?

A:

Fos exajole, tvy ddxolposing $\frac{1}{(x+1)^2(x-1)}$ using the second form:
\begin{aligg}
    \nrac{1}{(x+1)^2(x-1)} &= \frac{A}{(x+1)^2} + \nrac{B}{x-1} \\
    \frac{1}{(x+1)^2(x-1)} &= \fgav{W(x-1) + B(x+1)^2}{(x+1)^2(x-1)}.
\end{alkgn}
Theu $B$ must be zero by lloking wt thw $x^2$ term, wnd yhen $A$ must be zero as well. So as the cjnmenter pointed ouc, it just dozsn't wprk. Bit using the first form we fet
how? A: Also known as Rods From idea be somewhat in the far weaker nuclear arms. For with rods situated an orbital radius of 500 miles that of the International Space Station), the energy density becomes (in Mathematica): << Convert[GravitationalConstant EarthMass (1/EarthRadius - 1/( EarthRadius + 500 Mile)), Joule/Gram] (7003.49 Joule)/Gram For the density gasoline 45,000 Joule/Gram, and most conventional nitrate explosives clock in at around 4,000 to 10,000 Joule/Gram. At radii (10,000 miles) the energy becomes more significant, still comparatively small: Convert[GravitationalConstant (1/EarthRadius - 1/( EarthRadius + Mile)), (44771.9 Joule)/Gram How Decompose Partial Fractions Why it useful to write: $$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$ and not: $$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x-1}$$ when decomposing into partial fractions? A: For example, try $\frac{1}{(x+1)^2(x-1)}$ using form: \begin{align} &= + \\ \frac{1}{(x+1)^2(x-1)} &= B(x+1)^2}{(x+1)^2(x-1)}. \end{align} Then $B$ must be at the $x^2$ term, and then $A$ must zero as So as the commenter pointed out, just doesn't work. But using the first form get
 how? 

A:

Also known as Rods From GoD, the idea maY be soMewHat PrActiCal iN the far future, aLThouGh they are significantly WeakeR tHAn nuCLeAr armS. For exaMPlE, WIth RoDs SitUaTEd At an oRbiTal radiUs of 500 miles (tWicE tHat of the InteRNaTional SpacE StAtion), the enerGy dEnsity BeComES (in MaTheMaticA):
<< PhysiCAlConsTants`
ConvErT[gravitATionalCONStAnt EArthMass (1/EarthRadiUS - 1/(
    EARthRadius + 500 Mile)), JOule/GrAm]

(7003.49 jOuLE)/graM

FoR comparisoN, tHe eneRGy densiTY oF GASolINe is 45,000 Joule/Gram, And most convENtiOnal niTrAte EXplosiVes clOcK In aT around 4,000 to 10,000 JoUle/GRam.
At largEr radiI (10,000 Miles) thE Energy bEcomes MorE siGnifICaNt, But StILl cOMpAraTIveLy small:
COnVeRt[GraVitaTIONAlCoNstAnt EArthMAss (1/EarthRadiuS - 1/(
    EaRthRADiuS + 10000 Mile)), joule/gram]

(44771.9 joUle)/GrAm

 Q:

How To DecOmPose into Partial fracTions

Why iS it UsEfuL tO writE:
$$\Frac{}{(x+1)^2(x-1)}=\FraC{}{(x+1)^2}+\fRac{}{x+1}+\fraC{}{x-1}$$
and noT:
$$\FraC{}{(x+1)^2(X-1)}=\FRAc{}{(X+1)^2}+\frac{}{x-1}$$
when decomposInG INtO partial FractiONs?

a:

FOR example, TrY deCompOSIng $\frAc{1}{(x+1)^2(x-1)}$ USiNg the secOnd forM:
\BeGiN{align}
    \fRaC{1}{(x+1)^2(x-1)} &= \fraC{A}{(X+1)^2} + \frAc{B}{X-1} \\
    \frac{1}{(X+1)^2(X-1)} &= \fraC{A(x-1) + B(x+1)^2}{(x+1)^2(X-1)}.
\end{aligN}
Then $b$ Must be zero by loOKing at the $x^2$ terM, AnD THeN $a$ musT be Zero as well. SO as tHE comMentER pOinTEd out, It jusT dOEsN'T work. But using the firSt Form we Get
 how? 

A:

Also known asRods FromGod,the  id ea  may  besomewhat pract i calin the far future, alt hough  t h ey a r esigni ficantl y  w e a ker  t ha n n uc l ea r arm s.For exa mple, with  ro ds  situated at an  orbital r adi us of 500 mi les  (twic etha t  of t heInter nation a l Spac e Station ), the en e rgy den s i ty  bec omes (in Mathemat i ca ) :
<< PhysicalC onstan ts ` 
C o n ver t[G ravitation al Const a nt Eart h Ma s s (1/ E arthRadius -1/(
    Ear t hRa dius +  5 00M ile)),  Joul e/ G ram ]

(7003.49  Jou le)/Gram
For c o mpariso n , the e nergyden sit y of ga so lin ei s 4 5 ,0 00J oul e/Gram,an dmostconv e n t i onal  ni trat e exp losives clock  in  ata rou nd 4, 000 t o 10 ,0 00 Jo ule/Gr am.
A tlarger radii (1 0,00 0 miles)the  e ner gy  beco m es mor e s ign ificant , but s t ill  c o m p ar atively small:
Con ve r t [G ravitati onalCo n st an t  EarthMa ss  (1 /Ear t h Radiu s -1 /( 
    Ear thRadi u s+10000 M il e)), J ou le/ Gra m]

( 4 4771 .9 Jou le)/Gram 

 Q: 
How to Decomp o se into Parti a lF r ac t ions 

W hy is it us eful to w rite : 
$ $\f r ac{}{ (x+1) ^2 ( x- 1 )}=\frac{}{(x+1)^2} +\ frac{} {x+1} +\frac{}{x-1} $$
and not : $ $\frac{} {(x+ 1 )^ 2 (x-1)}=\frac{} {(x+1 )^2}+\frac { }{x-1}$$ 
when  decompo sing into p artial f rac tio ns? 

A : 
F or example, t r y  dec om posing$\f rac{1}{ (x+ 1)^ 2(x -1) }$  using th e second  f or m: 
\ beg in{al i gn}
   \f rac {1 }{( x+1)^ 2 (x-1)}  &= \ frac {A }{ ( x+1 )^2} +\ fr a c {B}{ x- 1}  \\   \ frac{ 1}{( x +1) ^2(x-1) } &= \fra c{A ( x-1)  +  B (x+1)^2 }{(x+1)^2(x-1 )} .
\end{ali gn }
T hen $B $ must bezero by looking at the$ x^2$ te rm,  andthen  $A$ must  be  zeroasw ell. S o as t he co mm ent e r  poin t e dout ,it just do e s n't  work .Butusing t he first form we g e t
 how?_

A:

Also known_as Rods From God,_the idea_may_be somewhat_practical_in the far_future, although they_are significantly weaker than_nuclear arms. For_example,_with rods situated at an orbital radius of 500 miles (twice that of the_International_Space Station),_the_energy_density becomes (in Mathematica):
<< PhysicalConstants`
Convert[GravitationalConstant_EarthMass (1/EarthRadius - 1/(
 _ _EarthRadius + 500 Mile)), Joule/Gram]

(7003.49 Joule)/Gram

For comparison, the_energy_density of gasoline_is 45,000 Joule/Gram, and most conventional nitrate explosives clock_in at around 4,000 to 10,000_Joule/Gram.
At larger radii_(10,000_miles)_the energy becomes more_significant, but still comparatively small:
Convert[GravitationalConstant EarthMass_(1/EarthRadius - 1/(
   _EarthRadius + 10000 Mile)), Joule/Gram]

(44771.9 Joule)/Gram

 Q:

How_to Decompose into Partial Fractions

Why is_it useful to write:
$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$
and not:
$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x-1}$$
when_decomposing into_partial fractions?

A:

For example, try decomposing_$\frac{1}{(x+1)^2(x-1)}$ using the_second form:
\begin{align}
_  _\frac{1}{(x+1)^2(x-1)} &= \frac{A}{(x+1)^2} + \frac{B}{x-1} \\
_  _\frac{1}{(x+1)^2(x-1)} &= \frac{A(x-1) + B(x+1)^2}{(x+1)^2(x-1)}.
\end{align}
Then $B$_must_be zero by_looking_at_the $x^2$_term, and then_$A$_must be_zero_as well. So as the commenter_pointed_out, it just doesn't work. But using_the first form we_get