Question 998860
{{{x}}}= height of the rectangular solid, in linear units.
"The width is 7 units more than the height" translates as
{{{x+7}}}= width of the rectangular solid, in linear units.
"The length is 1 unit more than eight times the height" translates as
{{{8x+1}}}= length of the rectangular solid, in linear units.
Since the volume is the product {{{height*width*length}}} and equals {{{750}}} cubic units,
{{{x(x+7)(8x+1)=750}}} is our equation.
It is cubic equation, and we could multiply and look for solutions whichever was we can.
However, should {{{x}}} be an integer, {{{x}}} , {{{(x+7)}}} and {{{(8x+1)}}} would be factors of {{{750}}} ,
with {{{x}}} being the smallest of the three, and {{{x+7}}} being {{{7}}} units more than {{{x}}}.
{{{750=10*75=(2*5)*(3*25)=2*3*5^3}}} .
Since the exponents of the prime factors are {{{1}}} , {{{1}}} , and {{{3}}} ,
{{{750}}} has {{{(1+1)*(1+1)*(3+1)=2*2*4=16}}} factors.
The smallest of them, in increasing order, are:
1, 2, 3, 5, 6, 10, 15, 25, and 30.
The only pair differing by {{{7}}} is {{{3}}} and {{{10}}} .
If {{{x=3}}} , {{{x+7=3+7=10}}} .
{{{750=10*75=10*3*25=3*10*25=x*(x+7)*(8x+1)}}}
Could it be that {{{system(x=3,x+7=10,8x+1=25)}}} ?
If {{{x=3}}} , {{{8x+1=8*3+1=24+1=25}}} , so {{{highlight(system(height=3,width=10,length=25))}}} (in linear units, of course).


THE CUBIC EQUATION:
Maybe you were expected to multiply and solve the resulting cubic equation.
That is a long and cumbersome process:
{{{x(x+7)(8x+1)=750}}}-->{{{x(8x^2+x+56x+7)=750}}}-->{{{x(8x^2+57x+7)=750}}}-->{{{8x^3+57x^2+7x=750}}}-->{{{8x^3+57x^2+7x-750=0}}}
The usual way to solve {{{8x^3+57x^2+7x-750=0}}} would be to find a rational solution, {{{p/q}}} ,
where {{{p}}} is a factor of {{{750}}} and {{{q}}} is a factor of {{{8}}} .
Luckily, {{{3}}} with {{{system {{{P=3,q=1}}} is a rational solution.
That means that the polynomial {{{8x^3+57x^2+7x-750}}} is divisible by {{{(x-3)}}} .
Dividing, we find
{{{(8x^3+57x^2+7x-750)/(x-3)=8x^2+81x+250)}}}<-->{{{8x^3+57x^2+7x-750=(x-3)*8x^2+81x+250)}}} .
So the solutions for {{{8x^3+57x^2+7x-750=0}}} are the solution for {{{x-3=0}}}<-->{{{x=3}}} ,
plus the solutions to {{{8x^2+81x+250=0}}} , if any.
Since {{{8x^2+81x+250=0}}} has no solutions, the only solution is {{{highlight(x=3)}}} .