Question

If $a=2,b=-2,$ find the value of:

1. $a^2+b^2$
2. $a^2+ab+b^2$
3. $a^2-b^2$

Answer 1

Step 1: Solve for the value of $a^2+b^2$

From the given details we get to know that $a=2,b=-2$

On substituting the values of$a$ and$b$ in the given algebraic equation and simplifying we get

$$\begin{gathered} ={\left(2\right)}^2+{(-2)}^2 \\ =4+4 \\ =8 \end{gathered}$$

Step 2: Solve for the value of $a^2+ab+b^2$

$$\begin{gathered} =2^2+(2\times -2)+{(-2)}^2 \\ =4+(-4)+\left(4\right) \\ =4–4+4 \\ =4 \end{gathered}$$

Step 3 :Solve for the value of $a^2–b^2$

$$\begin{gathered} =2^2–{(-2)}^2 \\ =4–\left(4\right) \\ =4–4 \\ =0 \end{gathered}$$

Hence, the required answer is,

1. $a^2+b^2=8$
2. $a^2+ab+b^2=4$
3. $a^2-b^2=0$