Divide $$ a^{3}+64 b^{3} $$ by $$ a+4 b $$ using long division

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$$
\textbf{Step 1: Write the Dividend in Standard Form}
$$

Write the dividend $$ a^3 + 64b^3 $$ as:
$$
a^3 + 0a^2 + 0a + 64b^3
$$

$$
\textbf{Step 2: Divide the Leading Term}
$$

Divide the leading term of the dividend, $$ a^3 $$, by the leading term of the divisor, $$ a $$:
$$
\frac{a^3}{a} = a^2
$$
This gives the first term of the quotient, $$ a^2 $$.

$$
\textbf{Step 3: Multiply and Subtract}
$$

Multiply the divisor $$ a+4b $$ by $$ a^2 $$:
$$
a^2 \cdot (a+4b) = a^3 + 4ba^2
$$
Subtract this from the dividend:
$$
\begin{array}{rcl}
\cline{1-1}
\quad & a^3 + 0a^2 + 0a + 64b^3 \
-(\, & a^3 + 4ba^2 & ) \
\cline{1-1}
\quad & 0a^3 - 4ba^2 + 0a + 64b^3 \
\end{array}
$$
The new remainder is:
$$
-4ba^2 + 0a + 64b^3
$$

$$
\textbf{Step 4: Repeat the Process}
$$

Divide the new leading term $$-4ba^2$$ by $$ a $$:
$$
\frac{-4ba^2}{a} = -4ab
$$
This is the next term of the quotient.

Multiply the divisor by $$-4ab$$:
$$
-4ab \cdot (a+4b) = -4ba^2 - 16ab^2
$$
Subtract this product from the current remainder:
$$
\begin{array}{rcl}
\cline{1-1}
\quad & -4ba^2 + 0a + 64b^3 \
-(\, & -4ba^2 - 16ab^2 & ) \
\cline{1-1}
\quad & 0a^2 + 16ab^2 + 64b^3 \
\end{array}
$$
The new remainder now is:
$$
16ab^2 + 64b^3
$$

$$
\textbf{Step 5: Continue the Division}
$$

Divide the leading term $$ 16ab^2 $$ by $$ a $$:
$$
\frac{16ab^2}{a} = 16b^2
$$
This is the next term of the quotient.

Multiply the divisor by $$ 16b^2 $$:
$$
16b^2 \cdot (a+4b) = 16ab^2 + 64b^3
$$
Subtract this product from the current remainder:
$$
\begin{array}{rcl}
\cline{1-1}
\quad & 16ab^2 + 64b^3 \
-(\, & 16ab^2 + 64b^3 & ) \
\cline{1-1}
\quad & 0 \
\end{array}
$$
The remainder is $$ 0 $$.

$$
\textbf{Step 6: State the Final Quotient}
$$

The quotient obtained is:
$$
a^2 - 4ab + 16b^2
$$

Thus, dividing $$ a^3+64b^3 $$ by $$ a+4b $$ gives the quotient:
$$
\boxed{a^2 - 4ab + 16b^2}
$$
The division of $$ a^3 + 64b^3 $$ by $$ a + 4b $$ results in $$ a^2 - 4ab + 16b^2 $$.

Divide \( a^{3}+64 b^{3} \) by \( a+4 b \) using long division

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