Question

solve in copy and explain
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Answer 1

$\large\textsf{Value of \sf{a^2-b^2-c^2-2ab-2bc-2ac} = 7}$

Step-by-step explanation :

• We are Asked to Find the value of $\sf{a^2-b^2-c^2-2ab-2bc-2ac}$ and also we are given with values of variables that are 2, -2, 1 for a, b, c respectively. To Solve for $\sf{a^2-b^2-c^2-2ab-2bc-2ac}$ just substitute the values of variables and Keep Solving the question and there we are done after the Doing Whole Answer. Don't Worry, I will give you all the steps to solve this

$\sf{a^2-b^2-c^2-2ab-2bc-2ac}$

• Substitute the value of a, b and c

$\to\sf{(2)^2-(-2)^2-(1)^2-2(2)(-2)-2(-2)(1)-2(2)(1)}$

$\to\sf{4-\left(-2\right)^2-\left(1\right)^2-2\left(2\right)\left(-2\right)-2\left(-2\right)\left(1\right)-2\left(2\right)\left(1\right)}$

$\to\sf{4-4-\left(1\right)^2-2\left(2\right)\left(-2\right)-2\left(-2\right)\left(1\right)-2\left(2\right)\left(1\right)}$

$\to\sf{4-4-1-2\left(2\right)\left(-2\right)-2\left(-2\right)\left(1\right)-2\left(2\right)\left(1\right)}$

$\to\sf{4-4-1-\left(-8\right)-2\left(-2\right)\left(1\right)-2\left(2\right)\left(1\right)}$

$\to\sf{4-4-1-\left(-8\right)-\left(-4\right)-2\left(2\right)\left(1\right)}$

$\to\sf{4-4-1-\left(-8\right)-\left(-4\right)-4}$

$\to\sf{0-1-\left(-8\right)-\left(-4\right)-4}$

$\to\sf{-1-\left(-8\right)-\left(-4\right)-4}$

$\to\sf{-1+8-\left(-4\right)-4}$

$\to\sf{7-\left(-4\right)-4}$

$\to\bf{7}$

Answer 2

$\large{\underline{\underline{\textsf{\maltese\: {\red{Required Answer :-}}}}}}$

$\odot \; {\underline{\bf{Given \; :}}}$

❖ a = 2

❖ b = -2

❖ c = 1

$\odot \; {\underline{\bf{Solution \; :}}}$

❖ a² - b² - c² - 2ab - 2bc - 2ca

= 2² - (-2)² - 1² - (2 × 2 × -2) - (2 × -2 × 1) - (2 × 1 × 2)

= 4 - (4) - 1 - (-8) - (-4) - 4

= 4 - 4 - 1 + 8 + 4 - 4

= 16 - 9

= 7