Question

What should be added to $a^6-64$ to be completely divisible by $a^4-16$?

Answer 1

Dividend $=a^6-64$

Divisor $=a^4-16$

Dividend, $a^6-64$ can be written as

$\Rightarrow {\left(a^2\right)}^3-{\left(2^2\right)}^3=(a^2-2^2)({\left(a^2\right)}^2+a^2{\left(2\right)}^2-{\left(2^2\right)}^2)$

$=(a^2-4)(a^4+4a^2+4^2)⟶\left(1\right)$

$\Rightarrow$ Divisor $=a^4-16={\left(a^2\right)}^2-{\left(4\right)}^2$

$=(a^2-4)(a^2+4)⟶\left(2\right)$

Now, dividing $\left(1\right)$ by $\left(2\right)$

$=\frac{(a^2-4)(a^4+4a^2+4^2)}{(a^2-4)(a^2+4)}$

Now, we are left with $\frac{(a^4+4a^2+4^2)}{(a^2+4)}$

Numerator can be written as ${(a^2+4)}^2-4a^2$

The first term is ${(a^2+4)}^2$ which is cleanly divisible by $(a^2+4).$ Only the second term is not divisible.

So, we can remove the second term by adding $4a^2$ in the numerator of $a^4+4a^2+4^2.$

So, effectively, we are adding $(a^2-4)\left(4a^2\right)$ in $a^6-64$ so that it is divisible by $a^4-16.$

The term to be added $=(a^2-4)4a^2=4a^2-16a^2.$

Hence, the answer is $4a^2-16a^2.$