If $x-y=3$ then ${x}^{3}-{y}^{3}=$ ?

Cierra Castillo
2022-07-02
Answered

If $x-y=3$ then ${x}^{3}-{y}^{3}=$ ?

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Dalton Lester

Answered 2022-07-03
Author has **12** answers

Step 1

We know that in general

${\text{XXX}}({x}^{3}-{y}^{3})=(x-y)({x}^{2}+xy+{y}^{2})$

and since

${\text{XXX}}x-y=3$

this gives us

${\text{XXX}}({x}^{3}-{y}^{3})=3({x}^{2}+xy+{y}^{2})$

There is no single solution.

The table below shows some of the possible combinations:

$\begin{array}{|ccc|}\hline \text{Given}\text{}x-y=3& & \text{In which case}\\ \text{if}\text{}x=& \text{then}\text{}y=& {x}^{3}-{y}^{3}=\\ -3& -6& 189\\ -2& -5& 117\\ -1& -4& 63\\ 0& -3& 27\\ 1& -2& 9\\ 2& -1& 9\\ 3& 0& 27\\ 4& 1& 63\\ 5& 2& 117\\ 6& 3& 189\\ \hline\end{array}$

We know that in general

${\text{XXX}}({x}^{3}-{y}^{3})=(x-y)({x}^{2}+xy+{y}^{2})$

and since

${\text{XXX}}x-y=3$

this gives us

${\text{XXX}}({x}^{3}-{y}^{3})=3({x}^{2}+xy+{y}^{2})$

There is no single solution.

The table below shows some of the possible combinations:

$\begin{array}{|ccc|}\hline \text{Given}\text{}x-y=3& & \text{In which case}\\ \text{if}\text{}x=& \text{then}\text{}y=& {x}^{3}-{y}^{3}=\\ -3& -6& 189\\ -2& -5& 117\\ -1& -4& 63\\ 0& -3& 27\\ 1& -2& 9\\ 2& -1& 9\\ 3& 0& 27\\ 4& 1& 63\\ 5& 2& 117\\ 6& 3& 189\\ \hline\end{array}$

Janessa Olson

Answered 2022-07-04
Author has **2** answers

Step 1

From equation (1) $y=x-3$

Substitution in (2) gives

3) ${x}^{3}-{(x-3)}^{3}=?$

Consider just the $(x-3)}^{3$

$(x-3)(x-3)(x-3)$

$(x-3)({x}^{2}-6x+9)$

${x}^{3}-9{x}^{2}+27x-27$

Substituting back into (3)

${x}^{3}-({x}^{3}-9{x}^{2}+27x-27)$

$9{x}^{2}-27x+27$

Set this to equal y

$\Rightarrow y=9{x}^{2}-27x+27$

Factor out the 9

4) $y=9({x}^{2}-3x+3)$

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 $(-\frac{1}{2})\times -3=+\frac{3}{2}$

${x}_{\text{vertex}}=+\frac{3}{2}$

By substitution in (4)

${y}_{\text{vertex}}=9[{\left(\frac{3}{2}\right)}^{2}-3\left(\frac{3}{2}\right)+3]=9\left[\frac{3}{4}\right]=\frac{27}{4}$

$\text{Vertex}\to (x,y)\to (\frac{3}{2},\frac{27}{4})$

From equation (1) $y=x-3$

Substitution in (2) gives

3) ${x}^{3}-{(x-3)}^{3}=?$

Consider just the $(x-3)}^{3$

$(x-3)(x-3)(x-3)$

$(x-3)({x}^{2}-6x+9)$

${x}^{3}-9{x}^{2}+27x-27$

Substituting back into (3)

${x}^{3}-({x}^{3}-9{x}^{2}+27x-27)$

$9{x}^{2}-27x+27$

Set this to equal y

$\Rightarrow y=9{x}^{2}-27x+27$

Factor out the 9

4) $y=9({x}^{2}-3x+3)$

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 $(-\frac{1}{2})\times -3=+\frac{3}{2}$

${x}_{\text{vertex}}=+\frac{3}{2}$

By substitution in (4)

${y}_{\text{vertex}}=9[{\left(\frac{3}{2}\right)}^{2}-3\left(\frac{3}{2}\right)+3]=9\left[\frac{3}{4}\right]=\frac{27}{4}$

$\text{Vertex}\to (x,y)\to (\frac{3}{2},\frac{27}{4})$

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