How do you evaluate ${(- 6 - - 6)}^{3} + 4$ $(- 6- - 6) ^ { 3} + 4$ ?

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Answer 1 · 2017-10-17T01:02:17

$4$ $4$

Focus on the parenthesis first. We have

$- 6 - - 6$ $-6--6$ which is simplified to $- 6 + 6$ $-6+6$ which is just $0$ $0$

So our expression is now

${(0)}^{3} + 4$ $(0)^3+4$

We now evaluate

$0^{3} + 4$ $0^3+4$

$= 0 + 4$ $=0+4$

$= 4$ $=4$

Answer 2 · 2017-10-17T01:03:05

The answer is $4$ $4$ .

Begin by simplifying within the parentheses.

${(- 6 - (- 6))}^{3} + 4 = 0^{3} + 4$ $(-6-(-6))^3+4 = 0^3+4$ because $- 6 - (- 6) = - 6 + 6 = 0$ $-6-(-6)=-6+6=0$

$0^{3} = 0$ $0^3=0$ , so $0^{3} + 4 = 4$ $0^3+4 = 4$

The answer is $4$ $4$ .

How do you evaluate (−6−−6)3+4(- 6- - 6) ^ { 3} + 4?