How do you evaluate $- 3 + 6 {(1 - 3)}^{2}$ $-3+6(1-3)^2$ ?

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Answer 1 · 2018-05-31T12:41:33

See a solution process below:

First, execute the Subtraction operation within the Parenthesis:

$- 3 + 6 {(1 - 3)}^{2} \Rightarrow - 3 + 6 {(- 2)}^{2}$ $-3 + 6(color(red)(1) - color(red)(3))^2 => -3 + 6(-2)^2$

Next, execute the Exponent operation:

$- 3 + 6 {(- 2)}^{2} \Rightarrow - 3 + 6 \times 4$ $-3 + 6(-2)^color(red)(2) => -3 + 6 xx 4$

Then, execute the Multiplication operation:

$- 3 + 6 \times 4 \Rightarrow - 3 + 24$ $-3 + color(red)(6) xx color(red)(4) => -3 + 24$

Now, execute the Additionoperation:

$- 3 + 24 \Rightarrow 21$ $color(red)(-3) + color(red)(24) => 21$

Answer 2 · 2018-06-06T09:13:44

$21$ $21$

Count the number of terms first and simplify each term to a single value. These are added or subtracted in the last line.

$- 3 + 6 {(1 - 3)}^{2}$ $color(blue)(-3)" + "color(green)(6(1-3)^2)$

Within each term do brackets first.
Then powers and roots
Then multiply and divide,

$= - 3 + 6 {(- 2)}^{2} \leftarrow$ $=color(blue)(-3)" + "color(green)(6(-2)^2)" "larr$ brackets

$= - 3 + 6 (4) \leftarrow$ $=color(blue)(-3)" + "color(green)(6(4)" "larr$ powers

$= - 3 + 24 \leftarrow$ $=color(blue)(-3)" + "color(green)(24)" "larr$ muliplication

$= 21 \leftarrow$ $= 21" "larr$ addition

How do you evaluate -3+6(1-3)^2−3+6(1−3)2-3+6(1-3)^2?