Réitigh (x+t)^3-x^3=3t(x+1)^2

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

Réitigh do x. (complex solution)

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

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} chun \left(x+t\right)^{3} a leathnú.

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

Comhcheangail x^{3} agus -x^{3} chun 0 a fháil.

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

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(x+1\right)^{2} a leathnú.

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

Úsáid an t-airí dáileach chun 3t a mhéadú faoi x^{2}+2x+1.

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

Bain 3tx^{2} ón dá thaobh.

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

Comhcheangail 3x^{2}t agus -3tx^{2} chun 0 a fháil.

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

Bain 6tx ón dá thaobh.

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

Bain t^{3} ón dá thaobh.

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

Comhcheangail na téarmaí ar fad ina bhfuil x.

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

Roinn an dá thaobh faoi 3t^{2}-6t.

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

Má roinntear é faoi 3t^{2}-6t cuirtear an iolrúchán faoi 3t^{2}-6t ar ceal.

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

Roinn t\left(3-t^{2}\right) faoi 3t^{2}-6t.

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

Réitigh do x.

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

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} chun \left(x+t\right)^{3} a leathnú.

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

Comhcheangail x^{3} agus -x^{3} chun 0 a fháil.

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

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(x+1\right)^{2} a leathnú.

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

Úsáid an t-airí dáileach chun 3t a mhéadú faoi x^{2}+2x+1.

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

Bain 3tx^{2} ón dá thaobh.

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

Comhcheangail 3x^{2}t agus -3tx^{2} chun 0 a fháil.

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

Bain 6tx ón dá thaobh.

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

Bain t^{3} ón dá thaobh.

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

Comhcheangail na téarmaí ar fad ina bhfuil x.

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

Roinn an dá thaobh faoi 3t^{2}-6t.

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

Má roinntear é faoi 3t^{2}-6t cuirtear an iolrúchán faoi 3t^{2}-6t ar ceal.

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

Roinn t\left(3-t^{2}\right) faoi 3t^{2}-6t.

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

Réitigh do t. (complex solution)

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

$$t=0$$

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

Réitigh do t.

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