Answer
Quotient: $\quad x^2+ax+a^2$
Remainder: $\quad0$
Work Step by Step
The given expression is:-
$(x^3-a^3)\div (x-a)$
Rewrite as descending powers of $x$.
$(x^3+0x^2+0x+a^3)\div (x-a)$
$\begin{matrix}
& x^2 & +ax &+a^2 & & \leftarrow &Quotient\\
&-- &-- &--&--& \\
x-a) &x^3&+0x^2&+0x&-a^3 & \\
&x^3 &-ax^2 & && \leftarrow &x^2(x-a) \\
& -- & -- & --& & \leftarrow &subtract \\
& 0 & ax^2 &+0x & \\
& & ax^2& -a^2x & & \leftarrow & ax(x-a) \\
& & -- & -- & -- & \leftarrow & subtract \\
& & 0& a^2x&-a^3 & \\
& & & a^2x& -a^3 & \leftarrow & a^2(x-a) \\
& & & -- & -- & \leftarrow & subtract \\
& & & 0&0 & \leftarrow & Remainder
\end{matrix}$
Checking:
(Quotient)(divisor)+ Remainder
$=(x^2+ax+a^2)(x-a)+0$
$=x^3+ax^2+a^2x-x^2a-a^2x-a^3$
$=x^3-a^3$
$=$ Dividend
Hence, the Quotient is $x^2+ax+a^2$.
and the remainder is $0$.
