Chapter 11.1, Problem 57CCR

To determine

To find: Simplify given expression.

Answer to Problem 57CCR

  ${\left(9+1\right)}^2+{\left(2-12\right)}^2=200$

Explanation of Solution

Given information:

  $6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2$

  $6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2$

On solving

We get

  $\begin{array}{l}6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2=6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2\\ \Rightarrow 6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2=6-1+{\left[3-\left(-2\right)\right]}^2\\ \Rightarrow 6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2=5+{\left[3+2\right]}^2\\ \Rightarrow 6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2=5+{\left[5\right]}^2\\ \Rightarrow 6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2=5+25\\ \Rightarrow 6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2=30\end{array}$

Therefore,

  $6-{\left(-1\right)}^2+{\left[3-\left(4-6\right)\right]}^2=30$