Question

a^3+b^3-3ab+1=0
solve this
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Answer 1

Answer:

ab³-4ab is the answer

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Answer 2

Answer:

There are a continuum of solutions to

a3+b3+3ab=1

Suppose that

x=a+b

then

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

which means

0=(x−1)(x2+(1−3a)x+3a2+1)

So either x=1 irregardless of a, or

x=3a−1±(a+1)−3−−−√2

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

That is, the only two real values of a+b are 1 and −2.

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