If a = _ x → 0 1 + 1 + x ^ 4 - 2 x ^ 4 and b = _ x → 0 ^ 2 x 2 - 1...

Given: $a = \lim_{x \to 0}\dfrac{\sqrt{1 + \sqrt{1 + {x}^{4}}} - \sqrt{2}}{{x}^{4}}$

$⇒ a = \lim_{x \to 0}\dfrac{\sqrt{1 + \sqrt{1 + {x}^{4}}} - \sqrt{2}}{{x}^{4}} \times \dfrac{\sqrt{1 + \sqrt{1 + {x}^{4}}} + \sqrt{2}}{\sqrt{1 + \sqrt{1 + {x}^{4}}} + \sqrt{2}}$

$⇒ a = \lim_{x \to 0}\dfrac{1 + \sqrt{1 + {x}^{4}} - 2}{{x}^{4}\left(\sqrt{1 + \sqrt{1 + {x}^{4}}} + \sqrt{2}\right)}$

$⇒ a = \lim_{x \to 0}\dfrac{\sqrt{1 + {x}^{4}} - 1}{{x}^{4}\left(\sqrt{1 + \sqrt{1 + 0}} + \sqrt{2}\right)}$

$⇒ a = \lim_{x \to 0}\dfrac{\sqrt{1 + {x}^{4}} - 1}{{x}^{4}\left(\sqrt{1 + 1} + \sqrt{2}\right)}$

$⇒ a = \dfrac{1}{2\sqrt{2}}\lim_{x \to 0}\dfrac{\sqrt{1 + {x}^{4}} - 1}{{x}^{4}}$

$⇒ a = \dfrac{1}{2\sqrt{2}}\lim_{x \to 0}\dfrac{\sqrt{1 + {x}^{4}} - 1}{{x}^{4}} \times \dfrac{\sqrt{1 + {x}^{4}} + 1}{\sqrt{1 + {x}^{4}} + 1}$

$⇒ a = \dfrac{1}{2\sqrt{2}}\lim_{x \to 0}\dfrac{1 + {x}^{4} - 1}{{x}^{4}\left(\sqrt{1 + {x}^{4}} + 1\right)}$

$⇒ a = \dfrac{1}{2\sqrt{2}}\lim_{x \to 0}\dfrac{{x}^{4}}{{x}^{4}\left(\sqrt{1 + 0} + 1\right)}$

$⇒ a = \dfrac{1}{2\sqrt{2}}\lim_{x \to 0}\dfrac{{x}^{4}}{{x}^{4}\left(2\right)}$

$⇒ a = \dfrac{1}{4\sqrt{2}}\lim_{x \to 0}\dfrac{{x}^{4}}{{x}^{4}}$

$⇒ a = \dfrac{1}{4\sqrt{2}}$

Also, $b = \lim_{x \to 0}\dfrac{\sin^{2}x}{\sqrt{2} - \sqrt{1 + \cos x}}$

$⇒ b = \lim_{x \to 0}\dfrac{\left(1 - \cos^{2}x\right) ( \sqrt{2} + \sqrt{1 + \cos x} )}{2 - ( 1 + \cos x )}$

$⇒ b = \lim_{x \to 0}\dfrac{\left(1 - \cos x\right)\left(1 + \cos x\right) ( \sqrt{2} + \sqrt{1 + \cos x} )}{\left(1 - \cos x\right)}$

$⇒ b = \lim_{x \to 0} ( 1 + \cos x ) ( \sqrt{2} + \sqrt{1 + \cos x} )$

$⇒ b = ( 1 + \cos 0 ) ( \sqrt{2} + \sqrt{1 + \cos 0} )$

$⇒ b = ( 1 + 1 ) ( \sqrt{2} + \sqrt{1 + 1} )$

$⇒ b = ( 2 ) ( 2\sqrt{2} )$

$⇒ b = 4\sqrt{2}$

$⇒ a{b}^{3} = \dfrac{1}{4\sqrt{2}} \times ( 4\sqrt{2}{)}^{3}$

$⇒ a{b}^{3} = 32$