If $$x = 1 + \sqrt{2} + \sqrt{3}$$ , then the value of $$(2x^4 - 8x^3 - 5x^2 + 26x- 28)$$ is __?

x = 1+ $$\sqrt {2} + \sqrt {3} $$
$$(x-1)^{2}$$ = $$(\sqrt {2} + \sqrt {3}) ^ {2} $$
$$x^{2} +1 - 2x = 5 + 2 \sqrt {6}$$
$$x^{2} - 2x = 4 + 2 \sqrt {6}$$ ( eq. (1) )
$$(x^{2} - 2x)^{2} = x^{4} + 4x^{2} - 4x^{3} = 40 + 16\sqrt{6} $$ eq (2)
Now in $$2x^{4} - 8x^{3} - 5x^{2} + 26x - 28 $$
or $$2(x^{4} - 4x^{3}) - 5x^{2} + 26x - 28 $$ ( putting values from eq (1) and eq (2) )
After solving we will get it reduced to $$6\sqrt{6}$$