Whakaoti i te (x^2+2x/sqrt{3}+1/3)(x^2-2x/sqrt{3}+1/3)

$$( x ^ { 2 } + \frac { 2 x } { \sqrt { 3 } } + \frac { 1 } { 3 } ) ( x ^ { 2 } - \frac { 2 x } { \sqrt { 3 } } + \frac { 1 } { 3 } )$$

Aromātai

$$\frac{\left(3x^{2}-1\right)^{2}}{9}$$

Ngā Hipanga Rongoā Poto

Whakangāwaritia te tauraro o \frac{2x}{\sqrt{3}} mā te whakarea i te taurunga me te tauraro ki te \sqrt{3}.

$$\left(x^{2}+\frac{2x\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+\frac{1}{3}\right)\left(x^{2}-\frac{2x}{\sqrt{3}}+\frac{1}{3}\right)$$

Ko te pūrua o \sqrt{3} ko 3.

$$\left(x^{2}+\frac{2x\sqrt{3}}{3}+\frac{1}{3}\right)\left(x^{2}-\frac{2x}{\sqrt{3}}+\frac{1}{3}\right)$$

Tā te mea he rite te tauraro o \frac{2x\sqrt{3}}{3} me \frac{1}{3}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)\left(x^{2}-\frac{2x}{\sqrt{3}}+\frac{1}{3}\right)$$

Whakangāwaritia te tauraro o \frac{2x}{\sqrt{3}} mā te whakarea i te taurunga me te tauraro ki te \sqrt{3}.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)\left(x^{2}-\frac{2x\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+\frac{1}{3}\right)$$

Ko te pūrua o \sqrt{3} ko 3.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)\left(x^{2}-\frac{2x\sqrt{3}}{3}+\frac{1}{3}\right)$$

Tā te mea he rite te tauraro o \frac{2x\sqrt{3}}{3} me \frac{1}{3}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)$$

Whakareatia te x^{2}+\frac{2x\sqrt{3}+1}{3} ki te x^{2}+\frac{2x\sqrt{3}+1}{3}, ka \left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)^{2}.

$$\left(x^{2}+\frac{2x\sqrt{3}+1}{3}\right)^{2}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x^{2} ki te \frac{3}{3}.

$$\left(\frac{3x^{2}}{3}+\frac{2x\sqrt{3}+1}{3}\right)^{2}$$

Tā te mea he rite te tauraro o \frac{3x^{2}}{3} me \frac{2x\sqrt{3}+1}{3}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\left(\frac{3x^{2}+2x\sqrt{3}+1}{3}\right)^{2}$$

Kia whakarewa i te \frac{3x^{2}+2x\sqrt{3}+1}{3} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

$$\frac{\left(3x^{2}+2x\sqrt{3}+1\right)^{2}}{3^{2}}$$

Pūrua 3x^{2}+2x\sqrt{3}+1.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+4\left(\sqrt{3}\right)^{2}x^{2}+6x^{2}+4\sqrt{3}x+1}{3^{2}}$$

Ko te pūrua o \sqrt{3} ko 3.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+4\times 3x^{2}+6x^{2}+4\sqrt{3}x+1}{3^{2}}$$

Whakareatia te 4 ki te 3, ka 12.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+12x^{2}+6x^{2}+4\sqrt{3}x+1}{3^{2}}$$

Pahekotia te 12x^{2} me 6x^{2}, ka 18x^{2}.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+18x^{2}+4\sqrt{3}x+1}{3^{2}}$$

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$\frac{9x^{4}+12\sqrt{3}x^{3}+18x^{2}+4\sqrt{3}x+1}{9}$$

Tauwehe

$$\frac{\left(3x^{2}-1\right)^{2}}{9}$$