(2x^3-4y^3)^2=dy eching

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

d uchun yechish (complex solution)

$$\left\{\begin{matrix}d=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}\text{, }&y\neq 0\\d\in \mathrm{C}\text{, }&x=0\text{ and }y=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2x^{3}-4y^{3}\right)^{2} kengaytirilishi uchun ishlating.

$$4\left(x^{3}\right)^{2}-16x^{3}y^{3}+16\left(y^{3}\right)^{2}=dy$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 3 va 2 ni ko‘paytirib, 6 ni oling.

$$4x^{6}-16x^{3}y^{3}+16\left(y^{3}\right)^{2}=dy$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 3 va 2 ni ko‘paytirib, 6 ni oling.

$$4x^{6}-16x^{3}y^{3}+16y^{6}=dy$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$dy=4x^{6}-16x^{3}y^{3}+16y^{6}$$

Tenglama standart shaklda.

$$yd=4x^{6}-16x^{3}y^{3}+16y^{6}$$

Ikki tarafini y ga bo‘ling.

$$\frac{yd}{y}=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}$$

y ga bo'lish y ga ko'paytirishni bekor qiladi.

$$d=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}$$

d uchun yechish

$$\left\{\begin{matrix}d=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}\text{, }&y\neq 0\\d\in \mathrm{R}\text{, }&x=0\text{ and }y=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2x^{3}-4y^{3}\right)^{2} kengaytirilishi uchun ishlating.

$$4\left(x^{3}\right)^{2}-16x^{3}y^{3}+16\left(y^{3}\right)^{2}=dy$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 3 va 2 ni ko‘paytirib, 6 ni oling.

$$4x^{6}-16x^{3}y^{3}+16\left(y^{3}\right)^{2}=dy$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 3 va 2 ni ko‘paytirib, 6 ni oling.

$$4x^{6}-16x^{3}y^{3}+16y^{6}=dy$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$dy=4x^{6}-16x^{3}y^{3}+16y^{6}$$

Tenglama standart shaklda.

$$yd=4x^{6}-16x^{3}y^{3}+16y^{6}$$

Ikki tarafini y ga bo‘ling.

$$\frac{yd}{y}=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}$$

y ga bo'lish y ga ko'paytirishni bekor qiladi.

$$d=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}$$

x uchun yechish (complex solution)

$$x\in \frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}+\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}e^{\frac{4\pi i}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}+\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}e^{\frac{2\pi i}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}+\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}e^{\frac{4\pi i}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}-\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}-\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}e^{\frac{2\pi i}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}-\sqrt{d}\right)}}{2}$$

x uchun yechish

$$\left\{\begin{matrix}x=\frac{2^{\frac{2}{3}}\sqrt[3]{4y^{3}-\sqrt{dy}}}{2}\text{; }x=\frac{2^{\frac{2}{3}}\sqrt[3]{4y^{3}+\sqrt{dy}}}{2}\text{, }&d\leq 0\text{ and }y\leq 0\\x=\frac{2^{\frac{2}{3}}\sqrt[6]{y}\sqrt[3]{4y^{\frac{5}{2}}-\sqrt{d}}}{2}\text{; }x=\frac{2^{\frac{2}{3}}\sqrt[6]{y}\sqrt[3]{4y^{\frac{5}{2}}+\sqrt{d}}}{2}\text{, }&y\geq 0\text{ and }d\geq 0\end{matrix}\right,$$