Sum of Reciprocals

Whenever we see something like $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$ , we try to manipulate it to make it easier to use. In this case, we can put everything in the denominator $abcd$ , as follows:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +\frac{1}{d}$

$\implies \frac{bcd}{abcd} + \frac{acd}{abcd} + \frac{abd}{abcd} + \frac{abc}{abcd}$

$\implies \frac{bcd + acd + abd + abc}{abcd}$

We know this because, we can take out like terms from each of the fractions (For example, $\frac{bcd}{abcd} = \frac{1}{a}$ , since the $bcd$ terms cancel out in both the numerator and the denominator.)

So now we know that we want to find the value of $\frac{abc + acd + abd + bcd}{abcd}$ . Notice however that the denominator, $abcd$ , is given to be $1$ in the question. So, all we need to find is the value of

$abc + acd+abd +bcd$

Now notice that the values of the terms we are given are

$a+b+c+d=0, abcd=1,\text{ and } a^3+b^3+c^3+d^3 = 1983$

Whenever I see something like this, I think to express one expression as a combination of the other two. Let's try to do that with this problem.

Since we see the third equation, $a^3+b^3+c^3+d^3$ , we can try cubing $a+b+c+d$ (With enough experience, this step is just intuition). Let's expand $(a+b+c+d)^3$ :

$(a+b+c+d)^3$

$\implies (a+b+c+d)(a+b+c+d)(a+b+c+d)$

$\implies a^3+b^3+c^3+d^3+3a^2b+3a^2c+3a^2d+3ab^2+3b^2c+3b^2d+3ac^2$ $+ 3bc^2+3c^2d+3ad^2+3bd^2+3cd^2+6abc+6acd+6abd+6bcd$

If you have trouble seeing this step, basically, we are taking each each term from each expansion of $a+b+c+d$ and multiplying it with each term in every other expansion. (Therefore, we can check to see if we expanded correctly if we have a total of $4^3=64$ terms in our expansion - here we are counting $6abc$ as 6 terms to be grouped together)

We can group our long expression into three smaller parts. We have

$a^3+b^3+c^3+d^3 =$

$a^3+b^3+c^3+d^3$

$+ 3a^2b+3a^2c+3a^2d+3ab^2+3b^2c+3b^2d+3ac^2+3bc^2+3c^2d+3ad^2+3bd^2+3cd^2$

$+6abc+6acd+6abd+6bcd$

Our first part, $a^3+b^3+c^3+d^3$ is easy to determine - it is given that this value is equal to 1983.

Our last part can be factored into $6(abc+acd+abd+bcd)$ That seems awfully familiar - it is equal to 6 times the value that we want to compute!

Our middle part is kind of tricky. First, we can take a factor of 3 out to obtain

$3(a^2b+a^2c+a^2d+ab^2+b^2c+b^2d+ac^2+bc^2+c^2d+ad^2+bd^2+cd^2)$

However, that still seems like a pain to compute. Luckily, we can factor this further by seperating the terms with $a^2$ , the terms with $b^2$ , and so on, to obtain

$3( a^2(b+c+d) + b^2(a+c+d) + c^2(a+b+d) + d^2(a+b+c)$

Let's take a look at our first term,

$a^2(b+c+d)$

Since we know that

$a+b+c+d = 0$

$b +c+d$ is simply

$a+b+c+d-a$

Or

$0-a = -a$

So the first term is just

$a^2 \cdot -a = -a^3$

Similarly, our second term, $b^2(a+c+d)$ , is just $-b^3$ , and so on.

So, our middle part can be simplified to

$3(-a^3-b^3-c^3-d^3)$ which is just

$-3(a^3+b^3+c^3+d^3)$

And since we know the value of $a^3+b^3+c^3+d^3$ , it can be simplified further to $-3 \times 1983$

We have simplified our long nasty expression into

$1983 + -3\times 1983 + 6(abc+acd+abd+bcd)$

Where $abc+acd+abd+bcd$ is what we want to compute. This can be further simplified to

$-2\times 1983 + 6(abc+acd+abd+bcd)$

However, notice that, since $a+b+c+d = 0$ , $(a+b+c+d)^3 = 0^3 = 0$

So we have the equation

$-2\times 1983 + abc+acd+abd+bcd = 0$

$\implies 6(abc+acd+abd+bcd) = 2\times 1983$

$\implies abc + acd + abd+bcd = \frac{2\times 1983}{6} = \frac{1983}{3} = \boxed{661}$ which is our desired answer. $\blacksquare$

$(a+b) = -(c+d)$

$a^3 + b^3 + c^3 + d^3 = 1983$

$\Rightarrow (a+b)((a+b)^2 - 3ab) + (c+d)((c+d)^2 - cd) = 1983$

$\Rightarrow -(c+d)((c+d)^2 - 3ab) + (c+d)((c+d)^2 - cd) = 1983$

$\Rightarrow 3(c+d)(ab-cd) = 1983 \Rightarrow (c+d)(ab-cd)=661$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$

= $\frac{ab(c+d) + bc(a+b)}{abcd}$

= $\frac{(c+d)(ab-cd)}{abcd} = \boxed{661}$

As $a+b+c+d=0 \Rightarrow a+b=-(c+d).$

As $a^3+b^3=(a+b)^3-3ab(a+b)=$ $-(c+d)^3+3ab(c+d)=-c^3-d^3-3cd(c+d)+3ab(c+d)$ $\Rightarrow a^3+b^3+c^3+d^3=3(c+d)(ab-cd)$ $\Rightarrow (c+d)(ab-cd)=\frac{1983}{3}=661.$

Now, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{a+b}{ab}+\frac{c+d}{cd} = -\frac{c+d}{ab}+\frac{c+d}{cd}=\frac{(c+d)(ab-cd)}{abcd}=(c+d)(ab-cd)=661$ So, Answer is $\boxed{661}$

Remember these and observe the pattern.

$(a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3b^{2}a$

$(a + b + c)^{3} = a^{3} + b^{3} + c^{3} + 3a^{2}b + 3a^{2}c + 3b^{2}a + 3b^{2}c + 3c^{2}a + 3c^{2}b + 6abc = a^{3} + b^{3} + c^{3} + 3a^{2}(b + c) + 3b^{2}(c + a) + 3c^{2}(a + b) + 6abc$

$(a + b + c + d)^{3} = a^{3} + b^{3} + c^{3} + d^{3} + 3a^{2}(b + c + d) + 3b^{2}(c + d + a) + 3c^{2}(d + a + b) + 3d^{2}(a + b + c) + 6(abc + bcd + cda + dab)$

Substitute the values,

$0 = 1983 + 3a^{2}(-a) + 3b^{2}(-b) + 3c^{2}(-c) + 3d^{2}(-d) + 6(abc + bcd + cda + dab)$

$0 = 1983 - 3a^{3} - 3b^{3} - 3c^{3} - 3d^{3} + 6(abc + bcd + cda + dab)$

$0 = 1983 - 3(a^{3} + b^{3} + c^{3} + d^{3}) + 6(abc + bcd + cda + dab)$

$0 = 1983 - 3(1983) + 6(abc + bcd + cda + dab)$

$abc + bcd + cda + dab = 661$ .......... (i)

We need to find $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$

$= \frac{abc + bcd + cda + dab}{abcd}$

$= 661$

That's the answer!

Rewrite the first equation as follows:

$\displaystyle a+b=-(c+d)$

Cube both the sides now to get:

$\displaystyle a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)$

$\displaystyle \Rightarrow a^3+b^3+c^3+d^3=-3(ab(a+b)+cd(c+d)$

Replace $a+b$ with $-(c+d)$ and $c+d$ with $-(a+b)$ .

$\displaystyle \Rightarrow a^3+b^3+c^3+d^3=3(ab(c+d)+cd(a+b))=3(abc+abd+cda+cdb)$

From the second equation, we have $abc=1/d, \,abd=1/c, \, cda=1/b$ and $cdb=1/a$ .

Hence,

$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1983}{3}=\fbox{661}$

$a+b+c+d=0$

$\Longrightarrow a+b=-(c+d)$

$\Longrightarrow (a+b)^3 = \left(-(c+d)\right)^3$

$\Longrightarrow a^3+b^3+3ab(a+b)=-\left(c^3+d^3+3cd(c+d)\right)$

$\Longrightarrow (a^3+b^3+c^3+d^3)+3\left(ab(a+b)+cd(c+d)\right)=0$

$\Longrightarrow ab(a+b)+cd(c+d) = -\dfrac{(a^3+b^3+c^3+d^3)}{3}$

$\Longrightarrow ab(a+b)+cd(c+d) = -\dfrac{1983}{3}=-661~~~ \cdot\cdot\cdot ~(*)$

Note that $a+b=-(c+d)\Longrightarrow c+d=-(a+b)$ ...

And, $~abcd=1 \Longrightarrow ab=\dfrac{1}{cd} \Longrightarrow cd = \dfrac{1}{ab}$ ...

Substituting these in $(*)$ , we get...

$\Longrightarrow ab(a+b)+cd(c+d) =-661$

$\Longrightarrow -\dfrac{c+d}{cd}+cd(c+d) =-661$

$\Longrightarrow \dfrac{c+d}{cd} = \dfrac{1}{c} + \dfrac{1}{d}=cd(c+d)+661~~~ \cdot\cdot\cdot ~ (**)$

Similarly, $\dfrac{1}{a} + \dfrac{1}{b}=ab(a+b)+661~~~ \cdot\cdot\cdot ~ (***)$

Summing up $(**)$ and $(***)$ , we get the following...

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} =ab(a+b)+661+cd(c+d)+661$

Note that...

$ab(a+b)+661+cd(c+d)+661$

$= 661+661+\left(ab(a+b)+cd(c+d)\right)$

$= 661+661-661=661$

Therefore, $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} = \fbox{661}$

First let's simplify $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ to get $abc+abd+acd+bcd$ .

Now let us expand $(a+b+c+d)^3=0$ to see what happens: $a^3+b^3+c^3+d^3+3(a^2b+a^2c+a^2d+b^2a+b^2c+b^2d+c^2a+c^2b+c^2d+d^2a+d^2b+d^2c)+6(abc+abd+acd+bcd)$ .

Note that since $b+c+d=-a$ , $a+c+d=-b$ , $a+b+d=-c$ and $b+c+d=-a$ we can simplify the previous expression to $a^3+b^3+c^3+d^3-3(a^3+b^3+c^3+d^3)+6(abc+abd+acd+bcd)$ which equals $6(abc+abd+acd+bcd)-2(a^3+b^3+c^3+d^3)=0$ .

We now substitute $a^3+b^3+c^3+d^3=1983$ to solve for $abc+abd+acd+bcd$ : $6(abc+abd+acd+bcd)-2(1983)=0\implies abc+abd+acd+bcd=\boxed{661}$

We have $\displaystyle \sum_{cyc}\frac{1}{a}=\displaystyle \sum_{cyc}(abc+abd)$ since $abcd=1$

$0=(a+b+c+d)^3=\displaystyle \sum_{cyc} a^3+ \frac{3!}{2!}\displaystyle \sum_{cyc} a^2(b+c+d) + (3!)\displaystyle \sum_{cyc} (abc+abd)$

Using $b+c+d=-a$ , we have

$0=\displaystyle \sum_{cyc} a^3 - 3 \displaystyle \sum_{cyc} a^3 + 6\displaystyle \sum_{cyc} (abc+abd)$

Hence $\displaystyle \sum_{cyc} \frac{1}{a}=\frac{1983}{3}=661$

Given,

$a+b+c+d=0$

multiply both side of the equation with $ab$ we get,

$a^{2}b+ab^{2}+abc+abd=0\Rightarrow\frac{1}{c}+\frac{1}{d}+a^{2}b+ab^{2}=0$ ..............(i)

[since, $abcd=1$ ]

similarly by multiplying ac,ad,bc,bd,cd we get,

$\frac{1}{b}+\frac{1}{d}+a^{2}c+ac^{2}=0$ ..............(ii)

$\frac{1}{b}+\frac{1}{c}+a^{2}d+ad^{2}=0$ ..............(iii)

$\frac{1}{a}+\frac{1}{d}+b^{2}c+bc^{2}=0$ ..............(iv)

$\frac{1}{a}+\frac{1}{c}+b^{2}d+bd^{2}=0$ ..............(v)

$\frac{1}{a}+\frac{1}{b}+c^{2}d+cd^{2}=0$ ..............(vi)

adding this equations (i) to (vi),

$3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})+a^{2}(b+c+d)+b^{2}(a+c+d)+c^{2}(a+b+d)+d^{2}(a+b+c)=0$

$\Rightarrow3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})+a^{2}(-a)+b^{2}(-b)+c^{2}(-c)+d^{2}(-d)=0$

$\Rightarrow3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})-(a^{3}+b^{3}+c^{3}+d^{3})=0$

$\Rightarrow3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=(a^{3}+b^{3}+c^{3}+d^{3})$

$\Rightarrow(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=\frac{1983}{3}=661$

hence,

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=661$

Let E represents the summation.

Given, E(a) = 0, ............................(i)

abcd =1, ........................(ii)

E(a^3) = 1983. ......................(iii)

If we solve the sum of reciprocals, we have to find E(abc) = ?

Now,

(a + b + c + d)^3 = E(a^3) + 3E((a^2) * (b + c + d)) + 6E(abc) ................(iv)

Now, putting values from (i),(ii),(iii) into (iv), we get

E(a^3) - 3E(a^3) + 6E(abc) = 0

3E(abc) = E(a^3)

E(abc) = 1983/3 = 661 (Ans)

Let $a+b=x$ , $ab=y$ . Then $c+d=-x$ and $cd=\frac{1}{y}$ .

What we want to find can be written as $\frac{a+b}{ab}+\frac{c+d}{cd}=\frac{x}{y}-xy$ after the substituting the numerators and denominators.

However, note that $a^{3}+b^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}-3ab(a+b)=x^{3}-3xy$ . Similarly, $c^{3}+d^{3}=-x^{3}+\frac{3x}{y}$ . Adding both of them, we find that the sum is $x^{3}-x^{3}-3xy+\frac{3x}{y}=-3(xy-\frac{x}{y})=3(\frac{x}{y}-xy)$ , which is 3 times what we want to find! Thus the answer is just $\frac{1983}{3}=661$ .

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = \frac{abc+abd+acd+bcd}{abcd} = abc+abd+acd+bcd$ . Since $abcd = 1$

Notice the following identity. It's motiavtion comes from:

$(a^3 + b^3 + c^3 - 3abc) = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$

$(a^{3}+b^{3}+c^{3}+d^{3}) - 3 (abc+abd+acd+bcd) = (a + b + c + d)(a^{2}+b^{2}+c^{2}+d^{2}-ab-ac-ad-bc-bd-cd)$

The RHS is $0$ since $a + b + c + d = 0$

$a^{3}+b^{3}+c^{3}+d^{3} - 3 (abc+abd+acd+bcd) = 0$

$1983 - 3 (abc+abd+acd+bcd) = 0 \Rightarrow abc + abd + acd + bcd = \boxed{661}$

Very very beatiful problem!!! :D (a+b+c+d)³=0 -> (a+b) = x and (c+d) = y. -> (x+y)³=0 -> x³+3x²y+3xy²+y³ = 0 -> x³+y³+3xy(x+y) = 0 Turn to x=a+b and y=c+d. (a+b)³+(c+d)³+3(a+b)(c+d)(a+b+c+d)=0. By a+b+c+d=0 identity: (a+b)³+(c+d)³=0 (a³+3ab(a+b)+b³)+(c³+3cd(c+d)+d³)=0 By a³+b³+c³+d³=1983 identity: 3ab(a+b)+3cd(c+d)+1983=0 Changing (a+b) = - (c+d) and (c+d)=-(a+b) by a+b+c+d=0 identity:

-3ab(c+d) - 3cd(a+b) = -1983 -3abc-3abd - 3acd - 3bcd = -1983 3abc+3abd+3acd+3bcd=1983 abc+abd+acd+bcd=661 Divides all by abcd. 1/a+1/b+1/c+1/d=661

Sorry for bad english.

(Key techniques: Substitution, Polynomial expansion.)

We shall refer to the three given equations, in order, simply as $(1), (2),$ and $(3)$ , respectively.

From $(2)$ , we can be sure that none of the variables is equal to zero. Hence, we can divide by any of the four variables (which we might need later).

Cubing $(1)$ , and rearranging terms, we have:

$a^3 + b^3 + c^3 + d^3 + 3a^2b + 3a^2c + 3a^2d + 3ab^2 + 3b^2c + 3b^2d + 3ac^2 + 3bc^2 + 3c^2d + 3ad^2 + 3bd^2 + 3cd^2 + 6abc + 6abd + 6acd + 6bcd = 0 (A)$

Quite long, isn't it? But we can let $(2)$ and $(3)$ enter the picture: from $(2)$ , the last four terms can be rewritten, with terms in reverse order, as $6 (\frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d})$ . And from $(3)$ , the first four terms can be replaced with 1983. Hence, equation $A$ , with the factor $3$ taken out of the terms after $1983$ , becomes:

$1983 + 3(a^2b + a^2c + a^2d + ab^2 + b^2c + b^2d + ac^2 + bc^2 + c^2d + ad^2 + bd^2 + cd^2) + 6 (\frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d}) = 0 (B)$

Now, we shall take note of the way by which the terms after 1983 are arranged. Let's take a look, for instance, on the first three terms: $a^2b + a^2c + a^2d$ . We can factor out $a^2$ , and the remaining three terms may look somewhat familiar: from $(1)$ , it is precisely $-a$ . Hence, the first three terms can be reduced to just $-a^3$ . A similar process can be done for the next succeeding groups of three terms, thus condensing $(B)$ into a prettier-looking equation:

$1983 + 3(-a^3 + (-b^3) + (-c^3) + (-d)^3) + 6(\frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d}) = 0 (C)$

Using equation $(3)$ again, we can reduce $(C)$ to just $1983 + 3(-1983) + 6(\frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d}) = 0 (D)$ . Transposition and division by 6 will give us the final answer:

$\frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d} = \boxed{661}$

Adding up the fraction gives us $\frac{cd(a+b)+ab(c+d)}{abcd}=\boxed{cd(a+b)+ab(c+d)}$

Use the identity $x^3+y^3=(x+y)^3-3xy(x+y)$

So $a^3+b^3+c^3+d^3=(a+b)^3-3ab(a+b)+(c+d)^3-3cd(c+d)=1983$

And since $c+d=-a-b$ , that means that when $(a+b)^3$ and $(c+d)^3$ are factored, they'll just cancel each other out to $0$

Therefore the above long expression and be simplified to $-3ab(a+b)-3cd(c+d)=1983$ and so $ab(a+b)+cd(c+d)=-661$

Keep in mind that

$a+b=-c-d$

$c+d=-a-b$

Multiply by $-1$ : $ab(-a-b)+cd(-c-d)=ab(c+d)+cd(a+b)=\boxed{661}$

c+d=-(a+b) cd=1/ab the given expression can be thus written as follow (a+b)/ab-ab(a+b)... ... and the cubic expression can thus be written as follow

(a+b)^3-3ab(a+b)-(a+b)^3+3(a+b)/ab=1983 or, (a+b)/ab-ab(a+b)=661.

1/a + 1/b + 1/c + 1/d = bcd + acd + abd + abc ................................... (1)

(a+b)³ = - (c+d)³

a³ + b³ + c³ + d³ = -3 [ ab (a + b) + cd (c + d) ]

but a + b = - (c + d)

Then

a³ + b³ + c³ + d³ = 3 [ ab (c + d) + cd (a + b) ]

a³ + b³ + c³ + d³ = 3 [bcd + acd + abd + abc ]

Then

bcd + acd + abd + abc = 1983/3 = 661 ..............................................(2)

From (1), (2)

1/a + 1/b + 1/c + 1/d = 661

we know that if x+y+z=0 then x^3+y^3+z^3=3xyz

so here we go ...let x=a, y=b , z=c+d by putting the identy we get

a^3+b^3+c^3+d^3+3cd(c+d)=3ab(c+d) 1983-3cd(a+b)=3ab(c+d) 1983=3(acd+bcd+abc+abd) 661=acd+bcd+abc+abd by deviding and multiplying with "abcd" we will get the result .and remember abcd=1(given). 661=1/a+1/b+1/c+1/d

can be solved in just 4 lines, use newton sums and assume the equation .

on performing addition, we get

$\frac{bcd+acd+abd+abc}{abcd}$

as abcd=1 we have bcd+acd+abd+acd let it be x

and from $(a+b+c+d)^{3}$ formula we get

$a^{3}+b^{3}+c^{3}+d^{3}+6(x) -3(a^{3}+b^{3}+c^{3}+d^{3})=0$

1983-3(1983)+6x=0

x=661

a+b+c+d =0 => a+b = -(c+d) => a^3 + b^3 + 3ab(a+b) = -c^3 - d^3 - 3cd( c+d ) => a^3 + b^3 + c^3 + d^3 = 3( c+d )( ab - cd) => (c+d)(ab - cd) = 661 1/a + 1/b + 1/c + 1/d = bcd + acd + abd +abc = cd( a+b ) +ab( c+d ) =(c+d)(ab - cd) =661

see that (a+b+c+d)^3=a^3+b^3+c^3+d^3+3a^2(b+c+d)+3b^2(c+d+a)+3c^2(a+b+d)+3d^2(a+b+c)+6(abc+bcd+cda+bad) =(-2) (a^3+b^3+c^3+d^3)+6(1/a+1/b+1/c+1/d) =>1/a+1/b+1/c+1/d=2 1983/6=661

$\displaystyle (a+b+c+d)^3=\sum_{sym} a^3+3 \sum_{sym}ab^2 +6\sum_{sym}$

from $a+b+c+d$ =0

$(a+b)^3=-(c+d)^3$ ,

$a^3+b^3+c^3+d^3=-(3a^2b+3b^2a+3c^2d+3cd^2)$

Use other combine we get

$\displaystyle a^3+b^3+c^3+d^3=-\sum_{sym}ab^2$

So,

$\displaystyle a^3+b^3+c^3+d^3+3 \sum_{sym} ab^2+6 \sum_{sym} abc=0$

$\displaystyle 1983-3.1983+6\sum_{sym} abc=0$

$\displaystyle 6 \sum_{sym} abc =2.1983=3966$

$\displaystyle \sum_{sym} abc=661$

$\displaystyle \frac{\sum_{sym} abc}{abcd}=\sum_{sym} \frac {1}{a}=661$

So, the answer is 661