Algebra

Pretty easy problem in this case, so I will try my best to explain it.  

 

So we can choose any two binomials to multiply, and then multiply it to the third binomial. For the sake of this problem, let's first multiply \((1+i) \)with \((2-i)\). Distributing and expanding, we get that \((1+i)(2-i) = 2 - i + 2i - i^2\). Combining like terms and knowing that \(i^2=-1\), we get \(2+i+1 = 3+i\). 

 

Now, we multiply this by the third binomial, which coincidentally is also \(3+i\). \((3+i)(3+i) = (3+i)^2 = 8 + 6 i\)

 

The final answer is \(8 + 6 i\)

 

Thanks!