What is [3-(2-4)^2+1] ?

The solution to [3-(2-4)^2+1] is 0

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[3-(2-4)^2+1]

[3 - (2 - 4)^{2} + 1]

0

Solution steps

3 - (2 - 4)^{2} + 1

Follow the PEMDAS order of operations

Calculate within parentheses

(2 - 4) : - 2

2 - 4

2 - 4 = - 2

= - 2

= 3 - (- 2)^{2} + 1

Calculate exponents

(- 2)^{2} : 4

(- 2)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2} = 4

= 4

= 3 - 4 + 1

Add and subtract (left to right)

3 - 4 + 1 : 0

3 - 4 + 1

3 - 4 = - 1

= - 1 + 1

- 1 + 1 = 0

= 0

= 0

(- (4)^{3})^{2}

5 (2 + 16 \div 4)

12 \div (- 4) \div (- 6)

(- 8 + 3) \times (5 - 9)

18 - 9 \div 3 \cdot 4 + 1