What is x*y if (squareroot2 -1)^5=x*squareroot2+y?

The only way to solve this equation is to multiply (sqrt2
-1)^5 out

(sqrt2-1)(sqrt2-1) =
2+1-2sqrt2=3-2sqrt2

(3-2sqrt2)^2=(sqrt2-1)^4=(3-2sqrt2)(3-2sqrt2)=9+8-12sqrt2=17-12sqrt2

ok,
the exponent here is five, not four, so lets multiply again a
(sqrt2-1)

(17-12sqrt2)(sqrt2-1)=17sqrt2-24+12sqrt2-17=39sqrt2-41

by
the form in the equation xsqrt2+y

we could say x=29 and
y=41

another way is to one-stepped multiply the 5th
exponent out by using the binomial
theorem.

(x-y)^5=x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5

the
coefficents are determined by the pascal triangle

since y=1
here, the y-terms are all one, so just omit it(. The formula
becomes

x^5-5x^4+10x^3-10x^2+5x-1

substitue
sqrt2 in there

4sqrt2-5*4+20sqrt2-10*2+5sqrt2-1=29sqrt2 -
41

Thus, x=29, and
y=41

Therefore, the requested product x*y is
29*-41= -1189