What is -(-2)^3-3(-2)^2+4 ?

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

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-(-2)^3-3(-2)^2+4

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

0

Solution steps

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

Follow the PEMDAS order of operations

Calculate exponents

(- 2)^{3} : - 8

(- 2)^{3}

Apply exponent rule:

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

n

is odd

(- 2)^{3} = - 2^{3} = - 8

= - 8

= - (- 8) - 3 (- 2)^{2} + 4

Calculate exponents

(- 2)^{2} : 4

(- 2)^{2}

Apply exponent rule:

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

n

is even

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

= 4

= - (- 8) - 3 \cdot 4 + 4

Multiply and divide (left to right)

3 \cdot 4 : 12

3 \cdot 4

3 \cdot 4 = 12

= 12

= - (- 8) - 12 + 4

Add and subtract (left to right)

- (- 8) - 12 + 4 : 0

- (- 8) - 12 + 4

Apply rule

- (- a) = a

- (- 8) = 8

= 8 - 12 + 4

8 - 12 = - 4

= - 4 + 4

- 4 + 4 = 0

= 0

= 0

\frac{- 20 + 2 \cdot 3}{8 \div ( - 2 ) + 2}

- \frac{9}{2} \times (- 10) - 18

29 - (8 + 15) - 27

- 2 (3)^{2} - 5 (3)

2 \times 2 + 4 \div 2 + 3 - (1 \times 1) + 7