Question 1102502
<br>
The approach used by tutor josgarithmetic is in fact valid; she just didn't use it correctly.<br>
The volume of the box, the way she set up the problem, should have been
{{{6(x-12)^2}}}<br>
She mistakenly wrote it as
{{{x(x-12)^2}}}<br>
Finishing the problem using this approach....<br>
{{{6(x-12)^2 = 1536}}}
{{{x-12)^2 = 256}}}
{{{x-12 = 16}}}  [we don't need to consider negative values, since we are talking about measurements of a piece of tin]
{{{x = 28}}}<br>
So the length of a side of the piece of tin is 28cm; the area is 28^2=784 cm^2.