Solve [15-(2^2+3*2)+(12-3^2)]^0=6

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

Verify

Solution Steps

Calculate 2 to the power of 2 and get 4.

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

Multiply 3 and 2 to get 6.

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

Add 4 and 6 to get 10.

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

Subtract 10 from 15 to get 5.

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

Calculate 3 to the power of 2 and get 9.

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

Subtract 9 from 12 to get 3.

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

Add 5 and 3 to get 8.

$$8^{0}=6$$

Calculate 8 to the power of 0 and get 1.

$$1=6$$

Compare 1 and 6.

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