Given that $x<1$ and \[(\log_{10} x)^2 - \log_{10}(x^2) = 48,\]compute the value of \[(\log_{10}x)^3 - \log_{10}(x^3).\]
This problem involves logarithms, which are inverse functions of exponents. One property of logarithms is that $\log_b(a^c) = c\log_b(a)$ for any base $b$, positive exponent $c$, and positive argument $a$. I can use this property to rewrite the given equation as \[(\log_{10} x)^2 - 2\log_{10}(x) = 48.\] This looks like a quadratic equation in $\log_{10} x$, which I can solve by factoring or using the quadratic formula. To make it easier, I will let $u = \log_{10} x$ and write the equation as \[u^2 - 2u - 48 = 0.\] This factors as \[(u-8)(u+6) = 0.\] So $u$ is either 8 or -6, which means $\log_{10} x$ is either 8 or -6. But wait, I have to remember that $x<1$, which means $\log_{10} x$ must be negative. So the only possible value for $\log_{10} x$ is -6. Now I have to find the value of \[(\log_{10}x)^3 - \log_{10}(x^3).\] I can use the same property of logarithms as before, and write this as \[(-6)^3 - 3\log_{10}(x).\] But I already know that $\log_{10} x = -6$, so I can plug that in and simplify. I get \[(-6)^3 - 3(-6) = -216 + 18 = -198.\] 