(x+t)^3-x^3=3t(x+1)^2 eching

$$( x + t ) ^ { 3 } - x ^ { 3 } = 3 t ( x + 1 ) ^ { 2 }$$

x uchun yechish (complex solution)

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

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$x^{3}+3x^{2}t+3xt^{2}+t^{3}-x^{3}=3t\left(x+1\right)^{2}$$

0 ni olish uchun x^{3} va -x^{3} ni birlashtirish.

$$3x^{2}t+3xt^{2}+t^{3}=3t\left(x+1\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

$$3x^{2}t+3xt^{2}+t^{3}=3t\left(x^{2}+2x+1\right)$$

3t ga x^{2}+2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}t+3xt^{2}+t^{3}=3tx^{2}+6tx+3t$$

Ikkala tarafdan 3tx^{2} ni ayirish.

$$3x^{2}t+3xt^{2}+t^{3}-3tx^{2}=6tx+3t$$

0 ni olish uchun 3x^{2}t va -3tx^{2} ni birlashtirish.

$$3xt^{2}+t^{3}=6tx+3t$$

Ikkala tarafdan 6tx ni ayirish.

$$3xt^{2}+t^{3}-6tx=3t$$

Ikkala tarafdan t^{3} ni ayirish.

$$3xt^{2}-6tx=3t-t^{3}$$

x'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(3t^{2}-6t\right)x=3t-t^{3}$$

Ikki tarafini 3t^{2}-6t ga bo‘ling.

$$\frac{\left(3t^{2}-6t\right)x}{3t^{2}-6t}=\frac{t\left(3-t^{2}\right)}{3t^{2}-6t}$$

3t^{2}-6t ga bo'lish 3t^{2}-6t ga ko'paytirishni bekor qiladi.

$$x=\frac{t\left(3-t^{2}\right)}{3t^{2}-6t}$$

t\left(3-t^{2}\right) ni 3t^{2}-6t ga bo'lish.

$$x=\frac{3-t^{2}}{3\left(t-2\right)}$$

x uchun yechish

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

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$x^{3}+3x^{2}t+3xt^{2}+t^{3}-x^{3}=3t\left(x+1\right)^{2}$$

0 ni olish uchun x^{3} va -x^{3} ni birlashtirish.

$$3x^{2}t+3xt^{2}+t^{3}=3t\left(x+1\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

$$3x^{2}t+3xt^{2}+t^{3}=3t\left(x^{2}+2x+1\right)$$

3t ga x^{2}+2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}t+3xt^{2}+t^{3}=3tx^{2}+6tx+3t$$

Ikkala tarafdan 3tx^{2} ni ayirish.

$$3x^{2}t+3xt^{2}+t^{3}-3tx^{2}=6tx+3t$$

0 ni olish uchun 3x^{2}t va -3tx^{2} ni birlashtirish.

$$3xt^{2}+t^{3}=6tx+3t$$

Ikkala tarafdan 6tx ni ayirish.

$$3xt^{2}+t^{3}-6tx=3t$$

Ikkala tarafdan t^{3} ni ayirish.

$$3xt^{2}-6tx=3t-t^{3}$$

x'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(3t^{2}-6t\right)x=3t-t^{3}$$

Ikki tarafini 3t^{2}-6t ga bo‘ling.

$$\frac{\left(3t^{2}-6t\right)x}{3t^{2}-6t}=\frac{t\left(3-t^{2}\right)}{3t^{2}-6t}$$

3t^{2}-6t ga bo'lish 3t^{2}-6t ga ko'paytirishni bekor qiladi.

$$x=\frac{t\left(3-t^{2}\right)}{3t^{2}-6t}$$

t\left(3-t^{2}\right) ni 3t^{2}-6t ga bo'lish.

$$x=\frac{3-t^{2}}{3\left(t-2\right)}$$

t uchun yechish (complex solution)

$$t=\frac{-\sqrt{9x^{2}+24x+12}-3x}{2}$$

$$t=0$$

$$t=\frac{\sqrt{9x^{2}+24x+12}-3x}{2}$$

t uchun yechish

$$\left\{\begin{matrix}\\t=0\text{, }&\text{unconditionally}\\t=\frac{\sqrt{9x^{2}+24x+12}-3x}{2}\text{; }t=\frac{-\sqrt{9x^{2}+24x+12}-3x}{2}\text{, }&x\geq -\frac{2}{3}\text{ or }x\leq -2\end{matrix}\right,$$