Question

1-3√5×1+3√5 solve this question.

Answer 1

a³+b³=(a+b)(a²+b²-ab)=(a+b){(a+b)²-3ab}

plug in :

a=(2+√5)^(1/3)

b=(2-√5)^(1/3)

ab={(2+√5)(2-√5)}^(1/3)=2²-5=-1

LET a+b=x

4=x(x²+3)

4=x³+3x

x³+3x-4=0

(x-1)(x²+x+4)=0

The above equation has ONE REAL root x=1 ,

the other two are complex ,since discriminant Δ =1²–4*1*4 < 0

∴ x=a+b=1

Answer 2

= 1-2.23+2.23.
=1. (-2.23+2.23=0)
the record, answer is 1