Question about finding cube root by hand

Question

To find the square root of $125$ by hand:

$$ \begin{array}{l|l} & 1,25.00 & 11.1 \\ &1 & \\ \hline 21 & \space \space \space \space 25 & \\ &\space \space \space \space21 & \\ \hline 221& \space \space \space \space \space \space 4.00& \\ &\space \space \space \space \space \space221& \end{array} $$

Thus the answer is $11.1$ and however many decimals one would desire. If you look at the method to calculate the square root, it quickly becomes obvious it is utilizing $(a+b)^2=a^2+2ab+b^2 = a^2+b(2a+b)$. The first digit of the solution is $1$, which is $a$, then this value is doubled and a $b$ is estimated, and finally multiplied by $b$ in the answer, or in other words $b(2a+b)$.

The cube root is something similar. Finding the cube root of $1278$ by hand:

$$ \begin{array}{l|l} & 1,278 & 1 \\ &1 & \\ \hline 300\times1^2=300 & \space \space \space \space 278 & \\ 30\times1\times b=?&\space \space \space \space & \\ b^2=?& & \\ \hline \end{array} $$

At this juncture obviously $300$ exceeds the partial dividend $278$, so the answer gets a $0$ and I bring down a new set of three digits, $000$:

$$ \begin{array}{l|l} & 1,278.000 & 10.8 \\ &1 & \\ \hline 300\times10^2=30000 & \space \space \space \space 278.000 & \\ 30\times10\times 8=2400&\space \space \space \space 259.712& \\ 8^2=64& & \\ \hline & \space \space \space \space \space \space \space \space 18288& \\ \end{array} $$

Thus the answer is $10.8$ and however many decimals one would desire. The cube root method utilizes $(10a+b)^3=1000a^3+300a^2b+30ab^2+b^3=1000a^3+b(300a^2+30ab+b^2)$. What I don't understand is why $10a$? Because intuitively I would think it should be $(a+b)^3$.

Answer 1

Hint: If you have a two digit number represented by $10a+b$, $b$ represents the units. What does $a$ represent?

(In general, a number can be represented by $10^a x_1 + 10^{a-1} x_2 + 10^{a-2} x_3 \dots 10^{a-(n-1)} x_{n-1} + 10^{(a-n)} x_n$, where $x_1, x_2, x_3, \dots x_{n-1}, x_n$ are the digits of that number.)