$c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Find the value of $a+b+c$

Question

$a=\sqrt{57+40\sqrt2}-\sqrt{57-40\sqrt2}$ and $b=\sqrt{25^{\frac{1}{\log_85}}+49^{\frac{1}{\log_67}}}$ and $c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Find the value of $a+b+c$.

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I tried to solve and simplify this problem but no luck.Please help me.Thanks.

Answer 1

$$a^2=57+40\sqrt2+57-40\sqrt2-2\sqrt{57^2-40^2\cdot2}=2\cdot57-2\sqrt49=100$$ $$a=10$$ $$b=\sqrt{5^{2\log_5 8}+7^{2\log_7 6}}=\sqrt{8^2+6^2}=\sqrt100=10$$ $$x=a-\frac1a,\quad a=(7+5\sqrt2)^{\frac13}$$ $$x^3+3x-14=\left(a^3-3a+\frac3a-\frac1{a^3}\right)+3\left(a-\frac1a\right)-14\\ =a^3-\frac1{a^3}-14=7+5\sqrt2-\frac1{7+5\sqrt2}-14=7+5\sqrt2-\frac{7-5\sqrt2}{-1}-14\\ =0=c$$

Answer 2

(1) $a=(5+4\sqrt{2})-(4\sqrt{2}-5)=10$

(2) $b=\sqrt{25^{\log_5{8}}+49^{\log_7{6}}}=\sqrt{8^2+6^2}=10$

(3) $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}=(\sqrt{2}+1)-\frac{1}{\sqrt{2}+1}=(\sqrt{2}+1)-(\sqrt{2}-1)=2$, $c=8+6-14=0$