Answer
$$9x^{2}-6x-4$$
Work Step by Step
1. First set out the question as a long division:
$(3x+2)\sqrt{ 27x^{3}+0x^{2}+0x-8}$
2. Next we work out how to get from '$3x$' to the first term '$27x^{3}$'... we would multiply $3x$ by $9x^{2}$ to get our first value $27x^{3}$
3. So, we write this above our equation so far so we don't forget it - like so
$9x^{2}$
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$(3x+2)\sqrt{ 27x^{3}+0x^{2}+0x-8}$
4. Now we multiply our $3x$ by $9x^{2}$ and write the result below our equation and we also multiply our two of $3x+2$ and write that underneath aswell:
$9x^{2}$
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$(3x+2)\sqrt{ 27x^{3}+0x^{2}+0x-8}$
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.subtract..... $27x^{3}+18x^{2}$
5. We now subtract the lower away section away from the upper section like so:
$(3x+2)\sqrt{ 27x^{3}+0x^{2}+0x-8}$
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.subtract..... $27x^{3}+18x^{2}$
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......................$0x^{3}-18x^{2}+0x-8$
6.We now want to find out how we get from $3x$ to '$-18x^{2}$' - we multiply $3x$ by $-6x$..
So we write this with our $9x^{2}$ like so:
$$9x^{2}-6x$$
7 . We multiply both $3x$ and $2$ by $-6x$ and write it underneath the reamining part of our equation like so:
$(3x+2)\sqrt{ 27x^{3}+0x^{2}+0x-8}$
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.subtract..... $27x^{3}+18x^{2}$
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......................$0x^{3}-18x^{2}+0x-8$
.subtract...............$-18x^{2}-12x$
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-leaving.............................$-12x-8$
8. Next, we divide our remainder by $3x+2$ like so:
$9x^2-6x+\frac{12x-8}{3x+2}$
where $\frac{12x}{3x+2}$ equals '$4+\frac{-16}{3x+2}$'
9. Lastly,we combine this with our $9x^{2}$ and $-6x$ off at the side to give us $9x^2-6x+4-\frac{16}{3x+2}$
10.$\frac{27x^{3}-8}{3x+2}=9x^2-6x+4-\frac{16}{3x+2}$
