What is the value of &(log 5)^3+(log 20)^3 & +(log 8)(log 0.25) where log denotes the base-ten logarithm?

2022 AMC 12A Problem 14

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14.

What is the value of

\begin{aligned} &(\log 5)^3+(\log 20)^3 \\ &\quad {}+(\log 8)(\log 0.25) \end{aligned}

where

\log

denotes the base-ten logarithm?

\dfrac32

\dfrac74

2

\dfrac94

3

Let

u=\log 2;

then

\log 5=1-u

and

\log 20=1+u

Also

\log 8=3u

and

\log 0.25=-2u;

use

a^3+b^3=(a+b)(a^2-ab+b^2)

Let

u=\log 2.

Then

\log 5=1-u,

\log 20=1+u,

\log 8=3u,

and

\log 0.25=-2u.

With

a=1-u,\ b=1+u,

we have

a+b=2

and

ab=1-u^2,

\begin{gathered} a^3+b^3 \\ =(a+b)\big((a+b)^2-3ab\big) \\ =2\big(4-3(1-u^2)\big) \\ =2+6u^2. \end{gathered}

The last term is

(3u)(-2u)=-6u^2,

so the total is

2+6u^2-6u^2=2.

Thus, the correct answer is C.