# If <mstyle displaystyle="true"> x - y = 3 </mstyle> then <msty

If $x-y=3$ then ${x}^{3}-{y}^{3}=$ ?
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Dalton Lester
Step 1
We know that in general
$\text{XXX}\left({x}^{3}-{y}^{3}\right)=\left(x-y\right)\left({x}^{2}+xy+{y}^{2}\right)$
and since
$\text{XXX}x-y=3$
this gives us
$\text{XXX}\left({x}^{3}-{y}^{3}\right)=3\left({x}^{2}+xy+{y}^{2}\right)$
There is no single solution.
The table below shows some of the possible combinations:
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Janessa Olson
Step 1
From equation (1) $y=x-3$
Substitution in (2) gives
3) ${x}^{3}-{\left(x-3\right)}^{3}=?$
Consider just the ${\left(x-3\right)}^{3}$
$\left(x-3\right)\left(x-3\right)\left(x-3\right)$
$\left(x-3\right)\left({x}^{2}-6x+9\right)$
${x}^{3}-9{x}^{2}+27x-27$
Substituting back into (3)
${x}^{3}-\left({x}^{3}-9{x}^{2}+27x-27\right)$
$9{x}^{2}-27x+27$
Set this to equal y
$⇒y=9{x}^{2}-27x+27$
Factor out the 9
4) $y=9\left({x}^{2}-3x+3\right)$
Step 2
From this point on it depends what you wish to do with it
Determine the vertex
Consider the $-3x$ in equation (4)
Apply $\left(-\frac{1}{2}\right)×-3=+\frac{3}{2}$
${x}_{\text{vertex}}=+\frac{3}{2}$
By substitution in (4)
${y}_{\text{vertex}}=9\left[{\left(\frac{3}{2}\right)}^{2}-3\left(\frac{3}{2}\right)+3\right]=9\left[\frac{3}{4}\right]=\frac{27}{4}$
$\text{Vertex}\to \left(x,y\right)\to \left(\frac{3}{2},\frac{27}{4}\right)$