<s>[INST] Roots of the equation $(1-4x)^4+32x^4=\frac{1}{27}​$ find all real roots of the equation $(1-4x)^4+32x^4=\dfrac{1}{27}​$
i try to use binomial expansion and am-gm inequality but i don't know how to do next. 
 [/INST] If $x<0$ so
$$(1-4x)^4+32x^4>1>\frac{1}{27}.$$
If $x>\frac{1}{4}$ so
$$(1-4x)^4+32x^4>32\left(\frac{1}{4}\right)^4>\frac{1}{27}.$$
Id est, it's enough to solve our equation for $x\in\left[0,\frac{1}{4}\right]$.
But in this case by Holder we obtain:
$$(1-4x)^4+32x^4=\frac{1}{27}(1+2)^3((1-4x)^4+2(2x)^4)\geq\frac{1}{27}(1-4x+2\cdot2x)^4=\frac{1}{27}.$$
But we have equality and the equality in Holder occurs for $(1,2)||((1-4x)^4,32x^4)$ only,  
which gives
$$1-4x=2x$$ or $$x=\frac{1}{6}.$$
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