Whakaoti i te (x-y)^3=A(x+y).text{provethat}(2x+y)dy/dx=x+2y

$$( x - y ) ^ { 3 } = A ( x + y ) . \text { prove that } ( 2 x + y ) \frac { d y } { d x } = x + 2 y$$

Whakaoti mō A

$$\left\{\begin{matrix}A=\frac{\left(x-y\right)^{3}}{x+y}\text{, }&x\neq -y\\A\in \mathrm{R}\text{, }&x=0\text{ and }y=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} hei whakaroha \left(x-y\right)^{3}.

$$x^{3}-3x^{2}y+3xy^{2}-y^{3}=A\left(x+y\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te A ki te x+y.

$$x^{3}-3x^{2}y+3xy^{2}-y^{3}=Ax+Ay$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$Ax+Ay=x^{3}-3x^{2}y+3xy^{2}-y^{3}$$

Pahekotia ngā kīanga tau katoa e whai ana i te A.

$$\left(x+y\right)A=x^{3}-3x^{2}y+3xy^{2}-y^{3}$$

He hanga arowhānui tō te whārite.

$$\left(x+y\right)A=x^{3}+3xy^{2}-y^{3}-3yx^{2}$$

Whakawehea ngā taha e rua ki te x+y.

$$\frac{\left(x+y\right)A}{x+y}=\frac{\left(x-y\right)^{3}}{x+y}$$

Mā te whakawehe ki te x+y ka wetekia te whakareanga ki te x+y.

$$A=\frac{\left(x-y\right)^{3}}{x+y}$$