Whakaoti i te [15-(2^2+3*2)+(12-3^2)]^0=6

$$[ 15 - ( 2 ^ { 2 } + 3 \cdot 2 ) + ( 12 - 3 ^ { 2 } ) ] ^ { 0 } = 6$$

Manatoko

Ngā Upane Otinga

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$\left(15-\left(4+3\times 2\right)+12-3^{2}\right)^{0}=6$$

Whakareatia te 3 ki te 2, ka 6.

$$\left(15-\left(4+6\right)+12-3^{2}\right)^{0}=6$$

Tāpirihia te 4 ki te 6, ka 10.

$$\left(15-10+12-3^{2}\right)^{0}=6$$

Tangohia te 10 i te 15, ka 5.

$$\left(5+12-3^{2}\right)^{0}=6$$

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

$$\left(5+12-9\right)^{0}=6$$

Tangohia te 9 i te 12, ka 3.

$$\left(5+3\right)^{0}=6$$

Tāpirihia te 5 ki te 3, ka 8.

$$8^{0}=6$$

Tātaihia te 8 mā te pū o 0, kia riro ko 1.

$$1=6$$

Whakatauritea te 1 me te 6.

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