150 days Streak problem

In general $(a + b)^{4} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4} = a^{4} + b^{4} + 4ab(a^{2} + b^{2}) + 6(ab)^{2}$

$\Longrightarrow (a + b)^{4} = a^{4} + b^{4} + 4ab((a + b)^{2} - 2ab) + 6(ab)^{2} = a^{4} + b^{4} + 4ab(a + b)^{2} - 2(ab)^{2}.$

Now with $\large a = \sqrt[4]{17 + 12\sqrt{2}}$ and $\large b = \sqrt[4]{17 - 12\sqrt{2}}$ we have that

$a^{4} + b^{4} = (17 + 12\sqrt{2}) + (17 - 12\sqrt{2}) = 34$ and $ab = \sqrt[4]{289 - 144*2} = 1.$

Thus $(a + b)^{4} = 34 + 4(a + b)^{2} - 2 \Longrightarrow (a + b)^{4} - 4(a + b)^{2} - 32 = 0$

$\Longrightarrow ((a + b)^{2} - 8)((a + b)^{2} + 4) = 0,$

so either $(a + b)^{2} = 8$ or $(a + b)^{2} = -4.$ Now $12\sqrt{2} \lt 12*1.4142 = 16.9704 \lt 17,$ so $b \gt 0,$ (as is $a$ ). Thus $a + b$ is real and positive, as will be $(a + b)^{2}.$ We can then conclude that $(a + b)^{2} = 8,$ and so $(a + b)^{4} = 8^{2} = \boxed{64}.$

$\begin{aligned} \text{Let } 17 + 12\sqrt{2} & = \left(a+b\sqrt{2}\right)^4 = a^4 + 4a^3b\sqrt{2} + 12a^2b^2 + 8ab^3\sqrt{2} + 4b^4 \end{aligned}$

Equating coefficients on both sides, we have:

$\begin{cases} a^4 + 12a^2b^2 + 4b^4 & = 17 \\ 4a^3b + 8ab^3 & = 12 \end{cases} \quad \Rightarrow a = 1 \text{ and } b = 1 \quad \Rightarrow \begin{cases} 17 + 12\sqrt{2} = \left(\sqrt{2}+1\right)^4 \\ 17 - 12\sqrt{2} = \left(\sqrt{2}-1\right)^4 \end{cases}$

Therefore,

$\begin{aligned} \left(\sqrt[4]{17 + 12\sqrt{2}} + \sqrt[4]{17 - 12\sqrt{2}}\right)^4 & = \left(\sqrt[4]{(\sqrt{2}+1)^4} + \sqrt[4]{(\sqrt{2}-1)^4}\right)^4 \\ & = \left(\sqrt{2} + 1 + \sqrt{2} - 1 \right)^4 \\ & = \left(2\sqrt{2}\right)^4 = \boxed{64} \end{aligned}$