Question

of a÷b+b÷a=-1(a,b not equal to zero) then value of a³-b³ is​

Answer 1

Step-by-step explanation:

x=ab  

Then you get:

x+1x=1

If x is positive, then this is always greater than 2, because:

x+1x=2+(x−−√−1x√)2

And similarly if x is negative this is always less than -2.

But there are solutions in complex numbers, you can rewrite as quadratic by multiplying both sides by x and rewriting in standard form.

x2−x+1=0

You will get:

x=1±i3√2

You can notice the fact that:

x=cos(π3)±isin(π3)

1x=cos(π3)∓isin(π3)

So for every integer k,

xk=cos(k×π3)±isin(k×π3)

Which means, whenever k is a multiple of 3 it is either -1 or +1.

x3m={−1+1 if m is odd if m is even

Now the original problem,

a3+b3=b3((ab)3+1)=b3(x3+1)=b3×0=0

More generally, For all integers m.

a6m+3+b6m+3=0

a6m−b6m=0