Question

solve a^3 + b^3 + 3ab = 1 find relation between a and b​

Answer 1

Answer:

There are a continuum of solutions to

a3+b3+3ab=1

a3+b3+3ab=1

Suppose that

x=a+b

x=a+b

then

1=a3+(x−a)3+3a(x−a)=x3−3ax2+3a2x+3ax−3a2

1=a3+(x−a)3+3a(x−a)=x3−3ax2+3a2x+3ax−3a2

which means

0=(x−1)(x2+(1−3a)x+3a2+1)

0=(x−1)(x2+(1−3a)x+3a2+1)

So either x=1x=1 irregardless of aa, or

x=3a−1±(a+1)−3−−−√2

x=3a−1±(a+1)−32

Thus, other than x=1x=1, the only real xx is −2−2, which comes from a=−1a=−1.

That is, the only two real values of a+ba+b are 11 and −2−2.