What is the result of $\lg^35+\lg^320+\lg8\lg0.25$?

Question

$$\lg^35+\lg^320+\lg8\lg0.25=$$

I tried this $$(\lg(5)+\lg(20))*[\lg^2(5)+\lg(2)*\lg(5) + \lg^2(2)]+3\lg(2)*(-2\lg(2))=$$

$$\lg(100)*[\lg^2(5)+\lg(2)*\lg(5) + \lg^2(2)]-6*\lg^2(2)=$$

Answer 1

$$\begin{align*}\log^3(5)+\log^3(20)+\log 8 \log 0.25 & = \log^3\left(\dfrac{10}{2}\right)+\log^3(10\cdot 2) + \log 2^3\log 2^{-2} \\ & = (1-\log 2)^3 + (1+\log 2)^3 -6(\log 2)^2 \\ & = 1-3\log 2+3\log^2 2-\log^3 2+1+3\log 2 + 3\log^2 2 + \log^3 2 - 6\log^2 2 \\ & = 2\end{align*}$$

Answer 2

Hint:$(\lg(5))^3+(\lg(20))^3=(\lg(5)+\lg(20)((\lg(5))^2-\lg(5)\lg(20)+(\lg(20))^2)$ and

$$\lg(5)+\lg(20)=\lg(100)=2$$ and $$\lg(8)=3\lg(2)$$ and $$\lg(0.25)=\lg(4^{-1})=\lg(2^{-2})=-2\lg(2)$$

Answer 3

$$(\log{5}+\log{20})^3-3(\log{20})(\log{5})(\log{5}+\log{20})+\log{8}\log{\frac{1}{4}}$$

$$(\log{\frac{10}{2}}+\log{10*2})^3-3(\log{10*2})(\log{\frac{10}{2}})(\log{\frac{10}{2}}+\log{10*2})+\log{2^3}\log{2^{-2}}$$

$$(2)^3-3((\log{10})^2-(\log{2})^2)(2)-6(\log{2})^2$$

$$8-6$$

$$2$$