Denominators are cubed

Applying Titu 's lemma :

$\dfrac{(\sin^2x)^2}{2}+\dfrac{(\cos^2x)^2}{3}\ge \dfrac{\left(\overbrace{\sin^2x+\cos^2x}^{\color{#D61F06}{1}}\right)^2}{2+3}=\dfrac1{5}$ So we see according to question equality holds in above inequality which is the case when $\dfrac{\sin^2x}2=\dfrac{\cos^2x}3$ or $\sin^2x =\dfrac{2}{5},\cos^2x=\dfrac3{5}$ . Direct substitution in required expression gives: $\dfrac{\left(\frac{2}{5}\right)^{4}}{8}+\dfrac{\left(\frac{3}{5}\right)^{4}}{27}=5^{-4}(2+3)=\dfrac1{125}$

$\huge \therefore 1+125=\boxed{\color{#007fff}{126}}$

$\LARGE\mathcal{\color{#007fff}{Alternate~Solution:-}}$

$\dfrac{(\sin^2 x)^2}{2}+\dfrac{(\cos^2 x)^2}{3}=\dfrac{1}{5}$ Substitute $\sin^2 x=t,\cos^2 x=1-\sin^2x=1-t$ ,

$\implies \dfrac{t^2}2+\dfrac{(1-t)^2}3=\dfrac 15$ $\implies 25t^2-20t+4=0\implies (5t-2)^2=0\implies t=\dfrac{2}5$ $\implies \sin^2 x=t=\dfrac 25, \cos^2 x=1-t=\dfrac 35$

• That's exactly what we have obtained earlier and rest is same.