Algebra problem solve for a,b,c and d?

Question

Can anyone find the values of these integers: a,b,c and d?

$$1+\sqrt{2}+\sqrt{3}+\sqrt{6} = \sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$

a+b+c+d = ?

Thank you.

Answer 1

$$F=1+\sqrt2+\sqrt3+\sqrt6=(1+\sqrt2)(1+\sqrt3)$$

$$F^2=a+\sqrt{b+\sqrt{c+\sqrt d}}$$

Squaring we get $$F^2=(1+\sqrt2)^2(1+\sqrt3)^2=(3+2\sqrt2)(4+2\sqrt3)=12+8\sqrt2+6\sqrt3+4\sqrt6\implies a=12$$

$$F^4=a^2+b+\sqrt{c+\sqrt d}+2a\sqrt{b+\sqrt{c+\sqrt d}}$$

$$F^4=(3+2\sqrt2)^2(4+2\sqrt3)^2=(17+12\sqrt2)(28+16\sqrt3)$$

$\implies a^2+b=17\cdot28$ but $a=12$

Can you take it home from here?